# On some counterparts of Rickart $*$-algebras

Arzikulov F. N.1,2 1 V.I. Romanovskiy Institute of Mathematics, Uzbekistan
Academy of Sciences, University 4, Olmazor, Tashkent, 100174, Uzbekistan,  2
Andijan State University, University 129, Andijan, 170100, Uzbekistan
<EMAIL_ADDRESS>and  Khakimov U. I.2<EMAIL_ADDRESS>
###### Abstract.

In the present paper, we introduce and study counterparts of Rickart
involutive algebras, i.e., almost inner Rickart algebras. We prove that a
nilpotent associative algebra, which has no nilpotent elements with nonzero
square roots, is an almost inner Rickart algebra. A nilpotent associative
algebra, which has no nilpotent elements with a square root $b$ such that
$b^{3}\neq 0$, is not an almost inner Rickart algebra if there exists a
nonzero element $a$ such that $a^{2}\neq 0$.

As a main result of the paper, we describe a finite-dimensional almost inner
Rickart algebra $\mathcal{A}$ over a field $\mathbb{F}$, isomorphic to
$\mathbb{F}^{n}\dot{+}\mathcal{N}$, $n=1,2$, with a nilradical $\mathcal{N}$.
Also, we classify finite-dimensional almost inner Rickart algebras over a
perfect field $\mathbb{F}$ with a nonzero nilradical $\mathcal{N}$.

###### Key words and phrases:

Associative algebra; nilradical; nilpotent associative algebra; nilpotent
element; idempotent

###### 2010 Mathematics Subject Classification:

16W10, 16E50, 16N40

Corresponding Author: Arzikulov F. N.

##  1\. Introduction

In the present paper, we introduce and study counterparts of Rickart
$*$-algebras, that is almost inner Rickart algebras. We define an almost inner
Rickart algebra as an associative algebra which is a Jordan algebra with
respect to the Jordan multiplication $a\cdot b=\frac{1}{2}(ab+ba)$ close to
inner RJ-algebras. Inner RJ-algebras are introduced and studied in the papers
[1], [2], [3].

The chosen notions were built around an (inner) quadratic annihilator. For
each nonempty subset $\mathcal{S}$ of an associative algebra $\mathcal{A}$,
the (inner) quadratic annihilator of $\mathcal{S}$ is defined by

${}^{\perp_{q}}\mathcal{S}:=\\{a\in\mathcal{A}:sas=0\forall
s\in\mathcal{S}\\}.$

Thus, following by [4], an associative algebra $\mathcal{A}$ which is equal to
$\mathcal{S}\dot{+}\mathcal{D}$, where $\mathcal{S}$ is a semisimple algebra
and $\mathcal{D}$ is a degenerate radical of $\mathcal{A}$ (including the case
$\mathcal{D}\equiv\\{0\\}$), is called an almost inner Rickart algebra if, for
each element $x\in\mathcal{A}$, there exists an idempotent $e\in\mathcal{A}$
such that
${}^{\perp_{q}}\\{x\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$,
where $\mathcal{S}^{2}:=\\{a^{2}:a\in\mathcal{S}\\}$,
$\mathcal{D}^{2}:=\\{a^{2}:a\in\mathcal{D}\\}$. There exist examples of almost
inner Rickart algebras without unit element (cf. [2, p. 32]). Note that, there
exist pairwise non-isomorphic (associative) almost inner Rickart algebras, the
Jordan algebras of which are isomorphic (see Examples 5)–8) below). This is a
motivation to introduce the notions of an almost inner Rickart algebra.

In the present paper, we prove that a nilpotent associative algebra, for every
element $a$ of which $a^{2}=0$, is an almost inner Rickart algebra. We explain
that a nilpotent associative algebra that has no nilpotent elements with a
square root $b$ such that $b^{3}\neq 0$ is not an almost inner Rickart algebra
if there exists a nonzero element $a$ such that $a^{2}\neq 0$.

As a main result of the paper we describe a finite-dimensional almost inner
Rickart algebra $\mathcal{A}$ over a field $\mathbb{F}$, isomorphic to
$\mathbb{F}^{n}\dot{+}\mathcal{N}$, $n=1,2$, with a nilradical $\mathcal{N}$
(Theorem 2.8). Also, we classify finite-dimensional almost inner Rickart
algebras over a perfect field $\mathbb{F}$ with a nonzero nilradical
$\mathcal{N}$.

##  2\. Description of finite-dimensional almost inner Rickart algebras

Throughout the paper let $\mathcal{A}$ be an associative algebra which is
equal to $\mathcal{S}\dot{+}\mathcal{D}$, where $\mathcal{S}$ is a semisimple
algebra and $\mathcal{D}$ is a degenerate radical of $\mathcal{A}$, and let

$\mathcal{S}^{2}:=\\{a^{2}:a\in\mathcal{S}\\},\mathcal{D}^{2}:=\\{a^{2}:a\in\mathcal{D}\\},$
${}^{\perp_{q}}\mathcal{S}:=\\{a\in\mathcal{A}:xax=0,(\forall
x\in\mathcal{S})\\}$

For an associative algebra $\mathcal{A}$, we consider the following condition:

(A) for every element $x\in\mathcal{A}$ , there exists an idempotent
$e\in\mathcal{A}$ such that

${}^{\perp_{q}}\\{x\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2}).$

Definition. An associative algebra satisfying condition (A) is called an
almost inner Rickart algebra, and condition (A) itself is called an almost
inner Rickart condition.

######  Lemma 2.1.

Let $\mathcal{A}=\mathcal{S}\dot{+}\mathcal{D}$ be an almost inner Rickart
algebra. Then the algebra $\mathcal{A}$ has an element $e$ such that
$ea^{2}=a^{2}e=a^{2}$ for every $a\in\mathcal{S}\cup\mathcal{D}$.

###### Proof.

Take $x=0$. We have
${}^{\perp_{q}}\\{x\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$
for some idempotent $e$ in $\mathcal{A}$. But
${}^{\perp_{q}}\\{x\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=\mathcal{A}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=\mathcal{S}^{2}\cup\mathcal{D}^{2}$.
Hence
$e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=\mathcal{S}^{2}\cup\mathcal{D}^{2}$.
Then $ea=e(ebe)=ebe=a$ and $ae=(ebe)e=ebe=a$ for all
$a\in\mathcal{S}^{2}\cup\mathcal{D}^{2}$. ∎

######  Lemma 2.2.

Let $\mathcal{A}=\mathcal{S}\dot{+}\mathcal{D}$ be an almost inner Rickart
algebra. Then the algebra $\mathcal{A}$ has no a nilpotent element $a$ such
that $a^{3}\neq 0$.

###### Proof.

Take an element $a\in\mathcal{A}$. Suppose that $a^{4}=0$. Then
$a^{2}\in{{}^{\perp_{q}}\\{a\\}}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$.
Since $\mathcal{A}$ is an almost inner Rickart algebra, there exists a nonzero
idempotent $e\in\mathcal{A}$ such that
${}^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$.
Therefore $aea=0$ and $ea^{2}=a^{2}$ since $e\in
e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$. Hence,
$a^{3}=aa^{2}=aea^{2}=(aea)a=0$.

Further, we shall apply the induction. The induction assumption is $a^{k}=0$,
$3\leq k\leq n$, implies that $a^{3}=0$, where the induction step is $n$. If
$n=4$ then the induction assumption is valid by the previous paragraph.

Suppose that $a^{n+1}=0$. Then, if $n=2k$, then $n\geq 4$ and
$a^{n+1}a=a^{2k+2}=(a^{2})^{k+1}=0$ and $1\leq k+1<n$. By the inductive
assumption $a^{3}=0$. So, $U_{a}a=0$ and
$a\in{{}^{\perp_{q}}\\{a\\}}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$. Since
$\mathcal{A}$ is an almost inner Rickart algebra, there exists an idempotent
$f\in\mathcal{A}$ such that
${}^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=f\mathcal{A}f\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$.
Therefore $fff=f\in f\mathcal{A}f\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$,
$afa=0$ and $fa=a$. Hence, $a^{2}=afa=0$ since $f\mathcal{A}f$ is a subalgebra
and $a^{2}\in f\mathcal{A}f$. By the previous paragraph $b^{3}=0$.

If $n=2k+1$, then $a^{n+1}=a^{2k+2}=(a^{2})^{k+1}=0$ and $1\leq k+1<n$. Again,
by the inductive assumption $a^{3}=0$. Hence, by the previous paragraph
$b^{3}=0$. Thus, by the induction, we obtain that for
$a\in(\mathcal{S}^{2}\cup\mathcal{D}^{2})$, if $a=b^{2}$ and $a^{n}=0$ for
some natural number $n$, then $b^{3}=0$. The proof is complete. ∎

######  Theorem 2.3.

A nilpotent associative algebra that has no nilpotent elements with a nonzero
square root is an almost inner Rickart algebra.

###### Proof.

Let $\mathcal{A}$ be a nilpotent associative algebra that has no nilpotent
elements with a nonzero square root. Let $a$ be a nonzero element from the
algebra $\mathcal{A}$. Since $\mathcal{A}$ is nilpotent, then $a^{m}=0$ for
some element $m$. Then $a^{2}=0$. Otherwise, the algebra $\mathcal{A}$
contains an element $b$, which is a nilpotent element with a nonzero square
root. So, $a^{2}=0$ for every non-zero element $a$ from $\mathcal{A}$.
Therefore, for any $a$, $b$ from $\mathcal{A}$,

$ab+ba=(a+b)^{2}-a^{2}-b^{2}=0,\,\,\,aba+baa=aba=0.$

The algebra $\mathcal{A}$ is an almost inner Rickart algebra. Indeed, the
algebra $\mathcal{A}$ does not contain a nonzero idempotent. In particular,
there is no non-zero idempotent $e$ such that

${}^{\perp_{q}}\\{{\bf 0}\\}=e\mathcal{A}e.$

However, since $(\mathcal{S}^{2}\cup\mathcal{D}^{2})=\\{{\bf 0}\\}$, we have

${}^{\perp_{q}}\\{{\bf 0}\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})={\bf
0}\mathcal{A}{\bf 0}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2}).$

Here ${}^{\perp_{q}}\\{{\bf 0}\\}=\mathcal{A}$. Similarly, for any element $a$
from $\mathcal{A}$, we have

${}^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=\\{{\bf
0}\\}={\bf 0}\mathcal{A}{\bf 0}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2}).$

The proof is complete. ∎

######  Theorem 2.4.

Let $\mathcal{A}$ be a nilpotent associative algebra. Then $\mathcal{A}$ is an
almost inner Rickart algebra if and only if, for any element $a$ in
$\mathcal{A}$, $a^{2}=0$.

###### Proof.

Suppose that $\mathcal{A}$ is an almost inner Rickart algebra. Since the
algebra $\mathcal{A}$ is nilpotent, there is no nonzero idempotent in the
algebra $A$. $A$ has only idempotent element ${\bf 0}$.

Let $a$ be an arbitrary nonzero element from the algebra $\mathcal{A}$ such
that $a^{2}\neq 0$. Since the algebra $\mathcal{A}$ is nilpotent, $a$ is a
nilpotent element. Then, by Lemma 2.2, $a^{3}=0$. Then

$a^{2}\in^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2}).$

Since the algebra $\mathcal{A}$ is nilpotent, there is no nonzero idempotent
in the algebra $A$. In particular, there is no nonzero idempotent $e$ such
that

$\\{a\\}^{\perp_{q}}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2}).$

So, $\mathcal{A}$ is not an almost inner Rickart associative algebra. This is
a contradiction. So, $a^{2}=0$.

Now, suppose that for any element $a$ in $\mathcal{A}$, $a^{2}=0$. Then, by
Theorem 2.3 $\mathcal{A}$ is an almost inner Rickart algebra. This ends the
proof. ∎

######  Theorem 2.5.

Let $\mathcal{A}$ be a finite-dimensional associative algebra with a one-
dimensional simple subalgebra $\mathcal{S}$ and an $n$-dimensional commutative
nilpotent radical $\mathcal{N}$ such that
$\mathcal{A}=\mathcal{S}\dot{+}\mathcal{N}$,
$\mathcal{N}^{2}=\\{a^{2}:a\in\mathcal{N}\\}\neq\\{0\\}$. Then there is a
nonzero idempotent $e\in\mathcal{A}$ such that
$\mathcal{A}=\mathbb{F}e\dot{+}\mathcal{N}$ and for any basis
$\\{e_{1},e_{2},\dots,e_{n}\\}$ of $\mathcal{N}$ the following conditions are
valid

$e_{i}e_{j}e_{k}=0,\\\ \\\ \ i,j,k=1,2,\ldots,n,$ $None$
$e(e_{i}e_{j})=(e_{i}e_{j})e=e_{i}e_{j},\ \\\ \ i,j=1,2,\ldots,n,$ $None$
$e_{i}ee_{j}+e_{j}ee_{i}=0,\\\ \\\ \ i,j=1,2,\ldots,n,$ $None$
$ee_{i}\in\mathcal{N},e_{i}e\in\mathcal{N}.$ $None$

Conversely, any associative algebra with a basis
$\\{e,e_{1},e_{2},\dots,e_{n}\\}$, where $e$ is an idempotent element,
satisfying the properties (1.1)–(1.4), is an almost inner Rickart algebra.

###### Proof.

By Lemma 2.2, the associative algebra $\mathcal{N}$ has no nilpotent elements
$a$ such that $a^{3}\neq 0$. Let $a$ be a nonzero element from the nilradical
$\mathcal{N}$. Then $a^{3}=0$. Therefore, for any $a,b$ from $\mathcal{N}$,

$0=(a+b)^{3}-a^{3}-b^{3}=(a^{2}+ab+ba+b^{2})(a+b)-a^{3}-b^{3},$
$=a^{3}+a^{2}b+aba+ab^{2}+ba^{2}+bab+b^{2}a+b^{3}-a^{3}-b^{3}=3a^{2}b+3b^{2}a,$
$0=(a-b)^{3}-a^{3}+b^{3}=(a^{2}-ab-ba+b^{2})(a-b)-a^{3}+b^{3},$
$=a^{3}-a^{2}b-aba+ab^{2}-ba^{2}+bab+b^{2}a-b^{3}-a^{3}-b^{3}=-3a^{2}b+3b^{2}a.$

Hence,

$a^{2}b=0,b^{2}a=0.$

As well as,

$abc=a\frac{1}{2}((b+c)^{2}-b^{2}-c^{2})=\frac{1}{2}(a(b+c)^{2}-ab^{2}-ac^{2})=0.$

So, for any elements $a,b,c$ from $\mathcal{N}$,

$abc=0.$ $None$

Let $a$ be an arbitrary nonzero element from the radical $\mathcal{N}$ such
that $a^{2}\neq 0$ and $\mathcal{N}^{2}=\\{x^{2}:x\in\mathcal{N}\\}$. Then
there exists a nonzero idempotent $e\in\mathcal{A}$ such that

$a^{2}\in^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}),\,\,\,aea=0.$

Due to (1.5) we have
$\mathcal{N}^{2}\subseteq^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$.
Hence,

$\mathcal{N}^{2}\subseteq
e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

As well as,

$a^{2}\in^{\perp_{q}}\\{a^{2}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Let $b$ be another nonzero element from the nilradical $\mathcal{N}$ such that
$b^{2}\neq 0$. Then there exists a nonzero idempotent $f\in\mathcal{A}$ such
that

$b^{2}\in^{\perp_{q}}\\{b\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=f\mathcal{A}f\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$

and

$\mathcal{N}^{2}\subseteq
f\mathcal{A}f\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$

Thus, $ea^{2}=fa^{2}=a^{2}$, $eb^{2}=fb^{2}=b^{2}$. Furthermore,

$ec=fc=c,\,\,\,c\in\mathcal{N}^{2}.$

Since codimension of $\mathcal{N}$ equals $1$ we have $e=f$. If we take $a+b$
then

$a^{2},\,\,b^{2},\,\,(a+b)^{2}\in^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=e\mathcal{A})e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}),$
$(a+b)^{2}=e(a+b)^{2}=a^{2}+2e(ab)+b^{2},$
$(a+b)^{2}=(a+b)^{2}e=a^{2}+2(ab)e+b^{2}.$

Hence,

$e(ab)=(ab)e=ab.$ $None$

Let ${e_{1},e_{2},\dots,e_{n}}$ be a basis in $\mathcal{N}$. Then, by (1.5) we
have (1.1). Also, by (1.6), we have (1.2). At the same time,

$e_{i}ee_{j}+e_{j}ee_{i}=0,\,\,i,j=1,2,\ldots,n,$

i.e. (1.3) is valid. Indeed, since $aea=0$, $a\in\mathcal{N}$, then, for any
$i$, $j$, $(e_{i}+e_{j})e(e_{i}+e_{j})=0$. Hence,

$e_{i}ee_{j}+e_{j}ee_{i}=0.$

So, (1.3) is valid. Since $\mathcal{N}$ is an ideal, we have (1.4) is also
valid.

Now let us prove that the associative algebra $\mathcal{A}$ with a basis
$\\{e,e_{1},e_{2},\dots,e_{n}\\}$, satisfying the properties (1.1)–(1.4), is
an almost inner Rickart algebra. We have

${}^{\perp_{q}}\\{e\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{{\bf
0}\\}={\bf 0}\mathcal{A}{\bf 0}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}),$

since $(\mathcal{S}^{2}\cup\mathcal{N}^{2})\subseteq e\mathcal{A}e$ by (1.2).

We take an arbitrary element of the form

$a=\sum_{i=1}^{n}\lambda_{i}e_{i}.$

By equalities (1.1) and (1.3), we have

$e\in^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=(\mathcal{S}^{2}\cup\mathcal{N}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Now we take an element of the form $\lambda e+a$, $\lambda\neq 0$. Since
$\mathcal{N}$ is an ideal, we have

$(\lambda e+a)b^{2}(\lambda e+a)=\lambda eb^{2}\lambda e+\lambda
eb^{2}a+ab^{2}\lambda e+ab^{2}a=\lambda eb^{2}\lambda e=\lambda^{2}b^{2}\neq
0$

for every $b\in\mathcal{N}$. We also have

$(\lambda e+a)(\mu e+b)^{2}(\lambda e+a)=(\lambda e+a)(\mu^{2}e+2\mu
eb+b^{2})(\lambda e+a)$ $=\lambda e\mu^{2}e\lambda e+c=\lambda^{2}\mu^{2}e+c,$

where

$c=2[\lambda e\mu^{2}ea+a\mu^{2}ea+a(\mu^{2}e+2\mu eb+b^{2})(\lambda e+a)]$

and $c\in\mathcal{N}$ since $\mathcal{N}$ is an ideal. Therefore $(\lambda
e+a)(\mu e+b)^{2}(\lambda e+a)\neq 0$. Hence,

${}^{\perp_{q}}\\{\lambda
e+a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0\\}={\bf 0}\mathcal{A}{\bf
0}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

So, $\mathcal{A}$ is an almost inner Rickart algebra. ∎

######  Remark 2.6.

Note that the associative algebra, indicated in theorem 2.5 is a direct sum of
an indecomposable non-nilpotent algebra and a nilpotent algebra. If, for
example, the nilradical $\mathcal{N}$ of an almost inner Rickart algebra
$\mathcal{A}$ is of codimension 2, then the quantity of non-nilpotent
indecomposable algebras in the direct sum of $\mathcal{A}$ may be one or two.
For example, the almost inner Rickart algebra

$\mathcal{A}_{1}:e_{1}^{2}=e_{1},e_{2}^{2}=e_{2},e_{1}n_{1}=n_{1},n_{2}e_{2}=n_{2},$

where $\\{e_{1},e_{2},n_{1},n_{2}\\}$ is a basis of the algebra
$\mathcal{A}_{1}$ and the omitted products vanish, is decomposable. The
subalgebras

$\mathcal{B}_{1}:e_{1}^{2}=e_{1},e_{1}n_{1}=n_{1},\mathcal{B}_{2}:e_{2}^{2}=e_{2},e_{2}n_{2}=n_{2}$

are almost inner Rickart algebras satisfying the conditions of theorem 2.5 and
$\mathcal{A}_{1}=\mathcal{B}_{1}\oplus\mathcal{B}_{2}$. The associative
algebra

$\mathcal{A}_{2}:e_{1}^{2}=e_{1},e_{2}^{2}=e_{2},n_{1}e_{1}=n_{1},e_{2}n_{1}=n_{1},n_{2}e_{1}=n_{2},$

where $\\{e_{1},e_{2},n_{1},n_{2}\\}$ is a basis of the algebra
$\mathcal{A}_{2}$ and the omitted products vanish, is indecomposable. The
nilradical $\mathcal{N}=<n_{1},n_{2}>$ is an almost inner Rickart algebra by
Theorem 2.4. Then $\mathcal{A}_{2}$ is also an almost inner Rickart algebra by
Theorem LABEL:aa. The subalgebras

$\mathcal{C}_{1}:e_{1}^{2}=e_{1},n_{1}e_{1}=n_{1},n_{2}e_{1}=n_{2},\mathcal{C}_{2}:e_{2}^{2}=e_{2},e_{2}n_{1}=n_{1}$

of the algebra $\mathcal{A}_{2}$ are almost inner Rickart algebras satisfying
the conditions of Theorem 2.5.

######  Theorem 2.7.

Let $\mathcal{A}$ be a finite-dimensional almost inner Rickart algebra over a
perfect field $\mathbb{F}$, with a nilpotent radical
$\mathcal{N}il(\mathcal{A})$. Then there exist pairwise orthogonal idempotents
$e_{1},e_{2},\dots,e_{m}$ in $\mathcal{A}$ such that

$\mathcal{A}=(e_{1}\mathbb{F}\oplus e_{2}\mathbb{F}\oplus\dots\oplus
e_{m}\mathbb{F})\dot{+}\mathcal{N}il(\mathcal{A}).$

###### Proof.

By the Cartan-Frobenius-Molien theorem, every finite-dimensional associative
algebra $\mathcal{A}$ over the real or complex numbers can be represented as
$\mathcal{A}=\mathcal{S}\dot{+}\mathcal{N}$, where $\mathcal{N}$ is a nil-
radical and $\mathcal{S}$ is a semi-simple algebra. In its turn $\mathcal{S}$
can be represented as

$\mathcal{S}=\mathcal{C}_{1}\oplus\mathcal{C}_{2}\oplus\dots\mathcal{C}_{n},$

where $\mathcal{C}_{i}$, $i=1,2,\dots n$, are simple algebras and
$\mathcal{C}_{i}\cong M_{n_{i}}(\mathcal{D}_{i})$, where
$M_{n_{i}}(\mathcal{D}_{i})$ is the algebra of $n_{i}\times n_{i}$ matrices
with entries from a division algebra $\mathcal{D}_{i}$. Therefore, by item 2)
of Remark 1.6 in [3], every finite-dimensional almost inner Rickart algebra
$\mathcal{A}$ has the following form

$\mathcal{A}=\mathcal{S}\dot{+}\mathcal{N}il(\mathcal{A}),$

where $\mathcal{N}il(\mathcal{A})$ is the nilpotent radical of $\mathcal{A}$
and $\mathcal{S}$ is a semisimple subalgebra every simple subalgebra of which
is isomorphic to only one of the following algebras: an abelian algebra, the
algebra of $2\times 2$-matrices over a division algebra, the algebra of
$3\times 3$-matrices over a division algebra.

Since the nilradical $\mathcal{N}il(\mathcal{A})$ does not contain nonzero
idempotents the factor-algebra $\mathcal{A}/\mathcal{N}il(\mathcal{A})$
contains all nonzero idempotents of the algebra $\mathcal{A}$. So, the factor-
algebra $\mathcal{A}/\mathcal{N}il(\mathcal{A})$ is an almost inner Rickart
algebra. Since $\mathcal{A}/\mathcal{N}il(\mathcal{A})\cong\mathcal{S}$ we
have $\mathcal{S}$ is also an almost inner Rickart algebra. At the same time,
$\mathcal{S}$ is a direct sum of simple algebras, every of which is an almost
inner Rickart algebra.

By the examples on page 35 of [2] and on page 8 of [3] the algebras
$M_{2}(\mathbb{F})$, $M_{3}(\mathbb{F})$, where
$\mathbb{F}=\mathbb{R},\mathbb{C},\mathbb{H}$, are not almost inner Rickart
algebras. Therefore, there exist pairwise orthogonal idempotents
$e_{1},e_{2},\dots,e_{m}$ in $\mathcal{A}$ such that

$\mathcal{S}=e_{1}\mathbb{F}\oplus e_{2}\mathbb{F}\oplus\dots\oplus
e_{m}\mathbb{F}.$

The proof is complete. ∎

######  Theorem 2.8.

Let $\mathcal{A}$ be an indecomposible finite-dimensional associative algebra
with a two-dimensional simple subalgebra $\mathcal{S}$ and an $n$-dimensional
commutative nilpotent radical $\mathcal{N}$ such that
$\mathcal{A}=\mathcal{S}\dot{+}\mathcal{N}$,
$\mathcal{N}^{2}=\\{a^{2}:a\in\mathcal{N}\\}\neq\\{0\\}$. Then there exist two
mutually orthogonal nonzero idempotents $p_{1}$, $p_{2}\in\mathcal{A}$ such
that $\mathcal{A}=(\mathbb{F}p_{1}\oplus\mathbb{F}p_{2})\dot{+}\mathcal{N}$
and for any basis $\\{e_{1},e_{2},\dots,e_{n}\\}$ of $\mathcal{N}$ one of the
following two cases is valid

the first case

$e_{i}e_{j}e_{k}=0,\\\ \\\ \ i,j,k=1,2,\ldots,n,$ $None$
$p_{1}(e_{i}e_{j})=(e_{i}e_{j})p_{1}=e_{i}e_{j},\ \\\ i,j=1,2,\ldots,n,$
$None$ $e_{i}p_{1}e_{j}+e_{j}p_{1}e_{i}=0,\\\ \\\ i,j=1,2,\ldots,n,$ $None$
$p_{k}e_{i}\in\mathcal{N},e_{i}p_{k}\in\mathcal{N},k=1,2,$ $None$
$p_{2}(e_{i}e_{j}+e_{j}e_{i})=0,i,j=1,2,\dots,n,$ $None$
$2p_{1}e_{i}p_{1}=p_{1}e_{i}+e_{i}p_{1},i=1,2,\dots,n,$ $None$

the second case

$e_{i}e_{j}e_{k}=0,\\\ \\\ \ i,j,k=1,2,\ldots,n,$ $None$
$e(e_{i}e_{j})=(e_{i}e_{j})e=e_{i}e_{j},\ \\\ \ i,j=1,2,\ldots,n,$ $None$
$e_{i}ee_{j}+e_{j}ee_{i}=0,\\\ \\\ \ i,j=1,2,\ldots,n,$ $None$
$ee_{i}\in\mathcal{N},e_{i}e\in\mathcal{N}.$ $None$ $\forall
b\in\mathcal{N}p_{1}b^{2}p_{1}\neq 0,p_{2}b^{2}p_{2}\neq 0\text{ if }b^{2}\neq
0.$ $None$

Conversely, any associative algebra with the basis
$\\{p_{1},p_{2},e_{1},e_{2},\dots,e_{n}\\}$, where $p_{1}$, $p_{2}$ are
idempotent elements, satisfying the properties (1.1)–(1.6) or the properties
(2.1)–(2.5), is an almost inner Rickart associative algebra.

###### Proof.

By Theorem 2.7, there exist pairwise orthogonal idempotents $p_{1},p_{2}$ in
$\mathcal{A}$ such that

$\mathcal{A}=(p_{1}\mathbb{F}\oplus p_{2}\mathbb{F})\dot{+}\mathcal{N}.$

By the proof of Theorem 2.5, we have, for any elements $a$, $b$, $c$ from
$\mathcal{N}$,

$a(bc)=(ab)c=0.$ $None$

Let $a$ be an arbitrary nonzero element from the radical $\mathcal{N}$ such
that $a^{2}\neq 0$ and $\mathcal{N}^{2}=\\{x^{2}:x\in\mathcal{N}\\}$. Then
there exists a nonzero idempotent $e\in\mathcal{A}$ such that

$a^{2}\in^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}),\,\,\,aea=0.$

Due to (1.1) we have
$\mathcal{N}^{2}\subseteq^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$.
Hence,

$\mathcal{N}^{2}\subseteq
e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

As well as,

$a^{2}\in^{\perp_{q}}\\{a^{2}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

There are following three possible cases

$\left\\{\begin{array}[]{ll}e=p_{1},&\hbox{}\\\ e=p_{2},&\hbox{}\\\
e=p_{1}+p_{2}.&\hbox{}\end{array}\right.$

Now, suppose that $e=p_{1}$, i.e.,

${{}^{\perp_{q}}\\{a\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=p_{1}\mathcal{A}p_{1}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

The conditions (1.2)–(1.4) can be found similar to the proof of Theorem 2.5.

We consider the annihilator ${{}^{\perp_{q}}\\{p_{2}\\}}$. It is clear that
$p_{1}\in{{}^{\perp_{q}}\\{p_{2}\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$
since $p_{1}\cdot p_{2}=0$ and $U_{p_{2}}p_{1}=0$. Also, from

$U_{p_{2}}(U_{p_{1}}(A))=U_{p_{2}p_{1}}(A)=0$

it follows that

$\mathcal{N}^{2}\subset
U_{p_{1}}(A)\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})\subset{{}^{\perp_{q}}\\{p_{2}\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Hence, $p_{2}a^{2}p_{2}=0$ for all $a\in N$. From
$p_{2}((a+b)^{2}-a^{2}-b^{2})p_{2}=0$ for all $a$, $b\in\mathcal{N}$ it
follows that $p_{2}(ab+ba)p_{2}=0$. Note that, since $A$ is an almost inner
Rickart algebra and
$p_{1}\in{{}^{\perp_{q}}\\{p_{2}\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$
we have

${{}^{\perp_{q}}\\{p_{2}\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{1}}(A)\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Now, we take the annihilator ${{}^{\perp_{q}}\\{p_{1}\\}}$. It is clear that
$p_{2}\in{{}^{\perp_{q}}\\{p_{1}\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$
since $p_{1}\cdot p_{2}=0$ and $U_{p_{1}}p_{2}=0$. Also, from

$\mathcal{N}^{2}\subset U_{p_{1}}(A))$

it follows that

$U_{p_{2}}(A)\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0,p_{2}\\}.$

Indeed, for any $a\in\mathcal{N}$,

$p_{2}(\lambda p_{1}+\mu p_{2}+a)^{2}p_{2}\neq(\lambda p_{1}+\mu p_{2}+a)^{2}$

if $\lambda\neq 0$ or $a\neq 0$. Hence,
$U_{p_{2}}(A)\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0,p_{2}\\}$. Thus,
$p_{2}(N^{2})=0$, $p_{2}a_{2}=0$ for any $a$, $b\in\mathcal{N}$. From this it
follows that $p_{2}(ab+ba)=0$. In particular,

$p_{2}(e_{i}e_{j}+e_{j}e_{i})=0,i,j=1,2,\dots,n.$ $None$

Note that, since $A$ is an almost inner Rickart algebra and
$p_{2}\in{{}^{\perp_{q}}\\{p_{1}\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$
we have

${{}^{\perp_{q}}\\{p_{1}\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{2}}(A)\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Note that

$p_{2}(\lambda p_{1}+a)^{2}p_{2}=0$

for any $a\in\mathcal{N}$, $\lambda\in{\mathbb{F}}$. Hence, $(\lambda
p_{1}+a)^{2}\in U_{p_{1}}(A)\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$, i.e.

$p_{1}(p_{1}a+ap_{1})p_{1}=p_{1}ap_{1}+p_{1}ap_{1}=2p_{1}ap_{1}=p_{1}a+ap_{1}.$

In particular,

$2p_{1}e_{i}p_{1}=p_{1}e_{i}+e_{i}p_{1},i=1,2,\dots,n.$ $None$

Now let us prove that an associative algebra $\mathcal{A}$ with a basis
$\\{p_{1},p_{2},e_{1},e_{2},\dots,e_{n}\\}$, satisfying the properties
(1.1)–(1.5), is an almost inner Rickart algebra.

We take the annihilator ${}^{\perp_{q}}\\{p_{1}\\}$. It is clear that
$p_{2}\in^{\perp_{q}}\\{p_{1}\\}$. We prove that

${}^{\perp_{q}}\\{p_{1}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{2}}(\mathcal{A})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Indeed, we have $U_{p_{2}}(\mathcal{A})\subseteq^{\perp_{q}}\\{p_{1}\\}$. The
equality

$p_{1}(\lambda p_{1}+\mu p_{2}+a)^{2}p_{1}=0,$

where $a\in\mathcal{N}$, is valid if $\lambda=0$. We have

$p_{1}(\mu p_{2}+a)^{2}p_{1}=p_{1}(\mu p_{2}^{2}+\mu p_{2}a+a\mu
p_{2}+a^{2})p_{1}=a^{2}\neq 0$

by (1.2). Hence,
${}^{\perp_{q}}\\{p_{1}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0,p_{2}\\}$.
At the same time, by (1.5), $p_{2}a^{2}=0$ for any $a\in\mathcal{N}$, and

$p_{2}(\lambda p_{1}+\mu p_{2}+a)^{2}p_{2}\neq(\lambda p_{1}+\mu p_{2}+a)^{2}$

if $\lambda\neq 0$ or $a\neq 0$. Hence,
$U_{p_{2}}(A)\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0,p_{2}\\}$. So,
${}^{\perp_{q}}\\{p_{1}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{2}}(\mathcal{A})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$.

Now we take the annihilator ${}^{\perp_{q}}\\{p_{2}\\}$. It is clear that
$p_{1}\in^{\perp_{q}}\\{p_{2}\\}$. We prove that

${}^{\perp_{q}}\\{p_{2}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{1}}(\mathcal{A})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Indeed, we have $U_{p_{1}}(\mathcal{A})\subseteq^{\perp_{q}}\\{p_{2}\\}$ and
$\mathcal{N}^{2}\subseteq U_{p_{1}}(\mathcal{A})$ by (1.2). Besides

$p_{1}(\mu p_{1}+a)^{2}p_{1}=p_{1}(\mu^{2}p_{1}^{2}+\mu p_{1}a+a\mu
p_{1}+a^{2})p_{1}=$ $\mu^{2}p_{1}+\mu
p_{1}(p_{1}a+ap_{1})p_{1}+a^{2}=\mu^{2}p_{1}+\mu 2p_{1}ap_{1}+a^{2}=$
$\mu^{2}p_{1}+\mu(p_{1}a+ap_{1})+a^{2}=(\mu p_{1}+a)^{2}$

by condition (1.6). Thus,

$U_{p_{1}}(\mathcal{A})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{(\mu
p_{1}+a)^{2}:a\in\mathcal{N}^{2},\lambda\in{\mathbb{F}}\\}.$

At the same time

$p_{2}(\mu p_{1}+a)^{2}p_{1}=0.$

Therefore,
${}^{\perp_{q}}\\{p_{2}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{1}}(\mathcal{A})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$.

Similar to the proof of theorem 2.5 we have, for any element of the form
$\lambda e+a$, $\lambda\neq 0$,

${}^{\perp_{q}}\\{\lambda
e+a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0\\}={\bf 0}\mathcal{A}{\bf
0}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Now we take an element of the form $\lambda p_{1}+\mu p_{2}+a$, $\lambda\neq
0$, $\mu\neq 0$ and consider the annihilator ${}^{\perp_{q}}\\{\lambda
p_{1}+\mu p_{2}+a\\}$. Since $\mathcal{N}$ is an ideal, by the properties of
the Peirce decomposition and (1.2), (1.5), we have

$(\lambda p_{1}+\mu p_{2}+a)b^{2}(\lambda p_{1}+\mu p_{2}+a)$ $=(\lambda
p_{1}+\mu p_{2})b^{2}(\lambda p_{1}+\mu p_{2})+(\lambda p_{1}+\mu
p_{2})b^{2}a+ab^{2}(\lambda p_{1}+\mu p_{2})+ab^{2}a$ $=(\lambda p_{1}+\mu
p_{2})b^{2}(\lambda p_{1}+\mu
p_{2})=\lambda^{2}p_{1}b^{2}p_{1}=\lambda^{2}b^{2}\neq 0$

if $b^{2}\neq 0$, for every $b\in\mathcal{N}$. By (1.5), We also have

$(\lambda p_{1}+\mu p_{2}+a)(\nu p_{1}+\rho p_{2}+b)^{2}(\lambda p_{1}+\mu
p_{2}+a)$ $=(\lambda p_{1}+\mu p_{2}+a)(\nu^{2}p_{1}+\rho^{2}p_{2}+\nu
bp_{1}+\rho bp_{2}+\nu p_{1}b+\rho p_{2}b+b^{2})^{2}(\lambda p_{1}+\mu
p_{2}+a)$ $=\lambda^{2}\nu^{2}p_{1}+\mu^{2}\rho^{2}p_{2}+c,$

where $c\in\mathcal{N}$ since $\mathcal{N}$ is an ideal. Therefore

$(\lambda p_{1}+\mu p_{2}+a)(\nu p_{1}+\rho p_{2}+b)^{2}(\lambda p_{1}+\mu
p_{2}e+a)\neq 0$

if $\nu\neq 0$ or $\rho\neq 0$. Hence,

${}^{\perp_{q}}\\{\lambda p_{1}+\mu
p_{2}+a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0\\}={\bf
0}\mathcal{A}{\bf 0}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Now we take an element of the form $\mu p_{2}+a$, $\mu\neq 0$ and consider the
annihilator ${}^{\perp_{q}}\\{\mu p_{2}+a\\}$. Since $\mathcal{N}$ is an
ideal, by the properties of the Peirce decomposition and (1.2), (1.5), we have

$(\mu p_{2}+a)b^{2}(\mu p_{2}+a)=\mu p_{2}b^{2}\mu p_{2}+\mu
p_{2}b^{2}a+ab^{2}\mu p_{2}+ab^{2}a$ $=\mu p_{2}b^{2}\mu p_{2}=0$

for every $b\in\mathcal{N}$.

By (1.3), (1.5), We also have

$(\mu p_{2}+a)(\nu p_{1}+\rho p_{2}+b)^{2}(\mu p_{2}+a)$ $=(\mu
p_{2}+a)(\nu^{2}p_{1}+\rho^{2}p_{2}+\nu bp_{1}+\rho bp_{2}+\nu p_{1}b+\rho
p_{2}b+b^{2})^{2}(\mu p_{2}+a)$ $=\mu^{2}\rho^{2}p_{2}+a\rho^{2}p_{2}a,$

if $\rho\neq 0$. Hence,

${}^{\perp_{q}}\\{\mu
p_{2}+a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0,p_{1},\mathcal{N}\\}.$

Since
$p_{1}\mathcal{A}p_{1}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0,p_{1},\mathcal{N}^{2}\\}$
by (1.2), we have

${}^{\perp_{q}}\\{\mu
p_{2}+a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=p_{1}\mathcal{A}p_{1}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

So, $\mathcal{A}$ is an almost inner Rickart algebra.

Now, suppose that $e=p_{1}+p_{2}$, i.e.,

${{}^{\perp_{q}}\\{a\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=(p_{1}+p_{2})\mathcal{A}(p_{1}+p_{2})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Then, similar to the proof of Theorem 2.5 we have the conditions (2.1)–(2.4)
are valid.

We have $\mathcal{N}^{2}\subseteq(p_{1}+p_{2})\mathcal{A}(p_{1}+p_{2})$.
Moreover, by (2.2)
$(\mathcal{S}^{2}\cup\mathcal{N}^{2})\subseteq(p_{1}+p_{2})\mathcal{A}(p_{1}+p_{2})$.
If, for example, there exists $b\in\mathcal{N}$ such that

${{}^{\perp_{q}}\\{b\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=p_{1}\mathcal{A}p_{1}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}),$

then we get the previous case. So we assume that, for every $b\in\mathcal{N}$,

${{}^{\perp_{q}}\\{b\\}}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=(p_{1}+p_{2})\mathcal{A}(p_{1}+p_{2})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Hence, for every $b\in\mathcal{N}$ such that $b^{2}\neq 0$, we have

$p_{1}b^{2}p_{1}\neq 0,p_{2}b^{2}p_{2}\neq 0.$ $None$

Now let us prove that an associative algebra $\mathcal{A}$ with a basis
$\\{p_{1},p_{2},e_{1},e_{2},\dots,e_{n}\\}$, satisfying the properties
(2.1)–(2.5), is an almost inner Rickart algebra.

We take an arbitrary element of the form

$a=\sum_{i=1}^{n}\lambda_{i}e_{i}.$

By equalities (2.1) and (2.3), we have

$e\in^{\perp_{q}}\\{a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=(\mathcal{S}^{2}\cup\mathcal{N}^{2})=e\mathcal{A}e\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

We take the annihilator ${}^{\perp_{q}}\\{p_{1}\\}$. It is clear that
$p_{2}\in^{\perp_{q}}\\{p_{1}\\}$. We prove that

${}^{\perp_{q}}\\{p_{1}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{2}}(\mathcal{A})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Indeed, by the condition (2.5)
$U_{p_{2}}(\mathcal{A})\cap\mathcal{N}^{2}=\oslash$. We have
$U_{p_{2}}(\mathcal{A})\subseteq^{\perp_{q}}\\{p_{1}\\}$. The equality

$p_{1}(\lambda p_{1}+\mu p_{2}+a)^{2}p_{1}=0,$

where $a\in\mathcal{N}$, is valid if $\lambda=0$. We have

$p_{1}(\mu p_{2}+a)^{2}p_{1}=p_{1}(\mu p_{2}^{2}+\mu p_{2}a+a\mu
p_{2}+a^{2})p_{1}=p_{1}a^{2}p_{1}\neq 0$

by (2.5). Hence,
${}^{\perp_{q}}\\{p_{1}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0,p_{2}\\}$.
At the same time, by (2.5), $p_{2}a^{2}p_{2}\neq 0$ for any $a\in\mathcal{N}$
such that $a^{2}\neq 0$, and

$p_{2}(\lambda p_{1}+\mu p_{2}+a)^{2}p_{2}=(\lambda p_{1}+\mu p_{2}+a)^{2}$

if $\lambda=0$ and $a=0$. Hence,
$U_{p_{2}}(A)\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0,p_{2}\\}$. So,
${}^{\perp_{q}}\\{p_{1}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{2}}(\mathcal{A})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})$.

Similarly, we have

${}^{\perp_{q}}\\{p_{2}\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=U_{p_{1}}(\mathcal{A})\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

Now we take an element of the form $\lambda p_{1}+\mu p_{2}+a$, $\lambda\neq
0$ or $\mu\neq 0$. Since $\mathcal{N}$ is an ideal, by the properties of the
Peirce decomposition, we have

$(\lambda p_{1}+\mu p_{2}+a)b^{2}(\lambda p_{1}+\mu p_{2}+a)$ $=(\lambda
p_{1}+\mu p_{2})b^{2}(\lambda p_{1}+\mu p_{2})+(\lambda p_{1}+\mu
p_{2})b^{2}a+ab^{2}(\lambda p_{1}+\mu p_{2})+ab^{2}a$ $=(\lambda p_{1}+\mu
p_{2})b^{2}(\lambda p_{1}+\mu p_{2})\neq 0$

for every $b\in\mathcal{N}$. We also have

$(\lambda p_{1}+\mu p_{2}+a)(\nu p_{1}+\rho p_{2}+b)^{2}(\lambda p_{1}+\mu
p_{2}+a)=$ $(\lambda p_{1}+\mu p_{2}+a)(\nu^{2}p_{1}+\rho^{2}p_{2}+\nu
bp_{1}+\rho bp_{2}+\nu p_{1}b+\rho p_{2}b+b^{2})^{2}(\lambda p_{1}+\mu
p_{2}+a)$ $=\lambda^{2}\nu^{2}p_{1}+\mu^{2}\rho^{2}p_{2}+c,$

where $c\in\mathcal{N}$ since $\mathcal{N}$ is an ideal. Therefore

$(\lambda p_{1}+\mu p_{2}+a)(\nu p_{1}+\rho p_{2}+b)^{2}(\lambda p_{1}+\mu
p_{2}+a)=\neq 0$

if $\nu\neq 0$ or $\rho\neq 0$. Hence,

${}^{\perp_{q}}\\{\lambda p_{1}+\mu
p_{2}+a\\}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2})=\\{0\\}={\bf
0}\mathcal{A}{\bf 0}\cap(\mathcal{S}^{2}\cup\mathcal{N}^{2}).$

So, $\mathcal{A}$ is an almost inner Rickart algebra. ∎

######  Proposition 2.9.

The associative algebra $M_{3}(\mathbb{C})$ does not have nilpotent elements
$a\in M_{3}(\mathbb{C})$ such that $a^{3}\neq 0$.

###### Proof.

Let

$M=\left(\begin{array}[]{ccc}a&b&c\\\ d&e&f\\\ g&h&j\\\ \end{array}\right)$

be a matrix from $M_{3}(\mathbb{C})$. Then

$M^{3}=(a+e+j)\cdot M^{2}+(bd+cg+fh-aj-ae-je)M+(bfg+cdh-cge-bdj-afh+aje)E.$

Let $\alpha=a+e+j$, $\beta=bd+cg+fh-aj-ae-je$ and $\gamma=bfg+cdh-cge-bdj-
afh+aje$. Then

${M}^{3}=\alpha\cdot M^{2}+\beta\cdot M+\gamma\cdot E,$

where $E$ is the unit matrix. Suppose that $M^{k}=0$, $M^{k-1}\neq 0$ for an
integer number $k>2$. Multiplying the last equality by $M^{k-1}$ on the right
side yields

$M^{k+2}=\alpha\cdot M^{k+1}+\beta\cdot M^{k}+\gamma\cdot M^{k-1}.$

Hence, $\gamma\cdot M^{k-1}=0$ and $\gamma=0$ since $M^{k-1}\neq 0$. So,

$M^{3}=\alpha\cdot M^{2}+\beta\cdot M.$

Multiplying the last equality by $M^{k-2}$ on the right side yields

$M^{k+1}=\alpha\cdot M^{k}+\beta\cdot M^{k-1}.$

From this, it follows that

$0=0+\beta\cdot M^{k-1}$

since $M^{k}=0$. At the same time $M^{k-1}\neq 0$. So, $\beta=0$. Similarly,
we have $\alpha=0$. Hence, $M^{3}=0$. This ends the proof. ∎

######  Remark 2.10.

If a finite-dimensional semisimple associative algebra is an almost inner
Rickart algebra, then this algebra is an inner RJ-algebra with respect to the
Jordan multiplication. By Proposition 2.9 and the example on page 8 of [3] the
condition of Lemma 2.2 is not sufficient for an associative algebra
$\mathcal{A}$ to be an almost inner Rickart algebra.

Examples. 1) It is obvious that every associative algebra which is an inner
RJ-algebra with respect to the multiplication $a\cdot b=\frac{1}{2}(ab+ba)$ is
an almost inner Rickart algebra.

2) Let $M_{n}({\mathbb{R}})$ be the associative algebra of $n\times n$
matrices over the field ${\mathbb{R}}$ of real numbers, where 1 is the unit of
the algebra $M_{n}({\mathbb{R}})$, ${\\{e_{i,j}\\}^{n}_{i,j=1}}$ is the system
of matrix units of the algebra $M_{n}({{\mathbb{R}}})$, $n=2k$.

a) Consider the associative algebra

$\mathcal{A}={\mathbb{R}}{\textbf{1}}+\sum_{i=1}^{k}{\mathbb{R}}e_{2i-1,2i}.$

This algebra is an almost inner Rickart algebra which is not involutive [1, p.
30].

b) Similarly, the associative algebra

$\mathcal{A}={\mathbb{R}}{\textbf{1}}+\sum_{i=1}^{k}{\mathbb{R}}e_{2i,2i-1},$

is an almost inner Rickart algebra.

c) Consider the associative algebra

$\mathcal{A}=\sum_{i=1}^{k}{\mathbb{R}}e_{2i-1,2i}$

This algebra is an almost inner Rickart algebra without a unit element [1,
Page 30, Example 3]. Indeed, since $(\mathcal{S}^{2}\cup\mathcal{D}^{2})={0}$,
then

${}^{\perp_{q}}\\{{\bf 0}\\}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})={\bf
0}(\mathcal{A}){\bf 0}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2}).$

Moreover,
$\\{e_{2i-1,2i}\\}^{\perp_{q}}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})={\bf
0}(\mathcal{A}){\bf 0}\cap(\mathcal{S}^{2}\cup\mathcal{D}^{2})$ for any $i$
from $\\{1,2,\dots,k\\}$.

d) Similar to example c) the associative algebra

$\mathcal{A}=\sum_{i=1}^{k}{\mathbb{R}}e_{2i,2i-1}$

is also an almost inner Rickart algebra without a unit element.

3) A vector space $\mathcal{A}$ of dimension $n+1$ with a basis
$\\{e,e_{1},e_{2},\dots,e_{n}\\}$ such that

$ee=e,\,\,ee_{i}=e_{i}e=e_{i},\,\,e_{i}^{2}=0,\,\,e_{i}e_{j}=0,i,j=1,2,...,n.$

is a commutative associative algebra and, by Theorem LABEL:a, an almost inner
Rickart algebra.

4) A vector space $\mathcal{A}_{2}^{2}$ of dimension $2$ with a basis
$\\{e_{1},e_{2}\\}$ such that

$e_{1}e_{1}=e_{1},e_{1}e_{2}=e_{2},e_{2}e_{1}=0,e_{2}e_{2}=0$

is an associative algebra (see Table 1) and the Jordan algebra
$(\mathcal{A}_{2}^{2},\cdot)$ is an inner RJ-algebra. At the same time a
vector space $\mathcal{A}_{2}^{3}$ of dimension $2$ with a basis
$\\{e_{1},e_{2}\\}$ such that

$e_{1}e_{1}=e_{1},e_{1}e_{2}=0,e_{2}e_{1}=e_{2},e_{2}e_{2}=0$

is an associative algebra (Table 1) and the Jordan algebra
$(\mathcal{A}_{2}^{3},\cdot)$ coincides with the inner RJ-algebra
$(\mathcal{A}_{2}^{2},\cdot)$.

5) A vector space $\mathcal{A}_{3}^{4}$ of dimension $3$ with a basis
$\\{e_{1},e_{2},e_{3}\\}$ such that

$e_{1}e_{3}=e_{2},e_{2}e_{3}=e_{2},e_{3}e_{3}=e_{3},$

the omitted products vanish, is an associative algebra (see Table 2) and the
Jordan algebra $(\mathcal{A}_{3}^{4},\cdot)$ is an inner RJ-algebra by Table
2. At the same time a vector space $\mathcal{A}_{3}^{6}$ of dimension $3$ with
a basis $\\{e_{1},e_{2},e_{3}\\}$ such that

$e_{3}e_{1}=e_{2},e_{3}e_{2}=e_{2},e_{3}e_{3}=e_{3},$

the omitted products vanish, is an associative algebra and the Jordan algebra
$(\mathcal{A}_{3}^{6},\cdot)$ coincides with the inner RJ-algebra
$(\mathcal{A}_{3}^{4},\cdot)$.

6) A vector space $\mathcal{A}_{4}^{8}$ of dimension $4$ with a basis
$\\{e_{1},e_{2},e_{3},e_{4}\\}$ such that

$e_{1}e_{1}=e_{1},e_{3}e_{1}=e_{3},e_{4}e_{3}=e_{2},e_{2}e_{1}=e_{2},$

the omitted products vanish, is an associative algebra (cf. Table 2.3 in [5])
and the Jordan algebra $(\mathcal{A}_{4}^{8},\cdot)$ is an inner RJ-algebra.
At the same time a vector space $\mathcal{A}_{4}^{9}$ of dimension $4$ with a
basis $\\{e_{1},e_{2},e_{3},e_{4}\\}$ such that

$e_{1}e_{1}=e_{1},e_{1}e_{3}=e_{3},e_{3}e_{4}=e_{2},e_{1}e_{2}=e_{2},$

the omitted products vanish, is an associative algebra (Table 2.3 in [5]) and
the Jordan algebra $(\mathcal{A}_{4}^{9},\cdot)$ coincides with the inner RJ-
algebra $(\mathcal{A}_{4}^{8},\cdot)$.

7) A vector space $\mathcal{A}_{4}^{19}$ of dimension $4$ with a basis
$\\{e_{1},e_{2},e_{3},e_{4}\\}$ such that

$e_{1}e_{1}=e_{1},e_{2}e_{2}=e_{2},e_{2}e_{3}=e_{3},e_{3}e_{1}=e_{3},e_{4}e_{1}=e_{4},$

the omitted products vanish, is an associative algebra (Table 2.3 in [5]) and
the Jordan algebra $(\mathcal{A}_{4}^{19},\cdot)$ is an inner RJ-algebra. At
the same time a vector space $\mathcal{A}_{4}^{20}$ of dimension $4$ with a
basis $\\{e_{1},e_{2},e_{3},e_{4}\\}$ such that

$e_{1}e_{1}=e_{1},e_{2}e_{2}=e_{2},e_{3}e_{2}=e_{3},e_{1}e_{3}=e_{3},e_{1}e_{4}=e_{4},$

the omitted products vanish, is an associative algebra (Table 2.3 in [5]) and
the Jordan algebra $(\mathcal{A}_{4}^{20},\cdot)$ coincides with the inner RJ-
algebra $(\mathcal{A}_{4}^{19},\cdot)$. Thus, two mutually non-isomorphic
almost inner Rickart algebras give the same inner RJ-algebra. The examples
above are a motivation to introduce the notion of an almost inner Rickart
algebra.

In the following tables Jordan algebras of dimension two and three are listed.
It is indicated that, wether a Jordan algebras is an almost inner Rickart
algebra or not for every Jordan algebra of the tables.

The fourth column of the tables indicates whether the appropriate Jordan
algebra is an almost inner Rickart algebra or not, i.e., if yes, then sign ”+”
is put in the appropriate place of the column, if not, then sign ”-” is put in
this place. All notations of the tables 1 and 2 are taken from [5].

Table 1

Two-dimensional associative algebras.

$J$ | Multiplication | Is this | According to
---|---|---|---
| table | algebra an | which theorem?
|  | almost inner Baer algebra? |
$\mathcal{A}_{2}^{1}$ | $n_{1}n_{1}=n_{2}$ | - | Theorem 2.4
$\mathcal{A}_{2}^{2}$ |  $e_{1}e_{1}=e_{1}$, | + | Theorem 2.5
| $e_{1}n_{1}=n_{1}$ |  |
$\mathcal{A}_{2}^{3}$ |  $e_{1}e_{1}=e_{1}$, | + | Theorem 2.5
| $n_{1}e_{1}=n_{1}$ |  |
$\mathcal{A}_{2}^{4}$ |  $e_{1}e_{1}=e_{1}$, | + | Theorem 2.5
|  $e_{1}n_{1}=n_{1}$, $n_{1}e_{1}=n_{1}$ |  |

Table 2

Three-dimensional associative algebras.

$J$ | Multiplication | Is this | According to
---|---|---|---
| table | algebra an inner | which theorem?
|  | Baer algebra? |
$\mathcal{A}_{3}^{1}$ |  $n_{1}n_{3}=n_{2}$, $n_{3}n_{1}=n_{2}$ | - | Theorem 2.4
$\mathcal{A}_{3}^{2}(\alpha)$ | $n_{1}n_{3}=n_{2}$ | - | Theorem 2.4
| $n_{3}n_{1}=\alpha n_{2},\alpha\in\mathbb{C}\setminus\\{1,-1\\}$ |  |
$\mathcal{A}_{3}^{2}(\alpha)$ | $n_{1}n_{3}=n_{2}$ | - | Theorem 2.4
| $n_{3}n_{1}=\alpha n_{2},\alpha=-1$ |  |
$\mathcal{A}_{3}^{3}$ |  $n_{1}n_{1}=n_{2}$, $n_{1}n_{2}=n_{3}$ | - | Theorem 2.4
| $n_{2}n_{1}=n_{3}$ |  |
$\mathcal{A}_{3}^{4}$ |  $n_{1}e_{1}=n_{2}$, $n_{2}e_{1}=n_{2}$ | + | Theorem 2.5
| $e_{1}e_{1}=e_{1}$ |  |
$\mathcal{A}_{3}^{5}$ |  $n_{2}e_{1}=n_{2}$, $e_{1}n_{1}=n_{1}$ | + | Theorem 2.5
| $e_{1}e_{1}=e_{1}$ |  |
$\mathcal{A}_{3}^{6}$ |  $e_{1}n_{1}=n_{2}$, $e_{1}n_{2}=n_{2}$ | + | Theorem 2.5
| $e_{1}e_{1}=e_{1}$ |  |
$\mathcal{A}_{3}^{7}$ |  $n_{1}e_{1}=n_{1}$, $e_{1}e_{1}=e_{1}$ | + | Theorem 2.8
|  $e_{2}n_{1}=n_{1}$, $e_{2}e_{2}=e_{2}$ |  |
$\mathcal{A}_{3}^{8}$ |  $n_{1}e_{1}=n_{1}$, $n_{2}e_{1}=n_{2}$ | + | Theorem 2.5
|  $e_{1}n_{1}=n_{1}$, $e_{1}e_{1}=e_{1}$ |  |
$\mathcal{A}_{3}^{9}$ |  $n_{2}e_{1}=n_{2}$, $e_{1}n_{1}=n_{1}$ | + | Theorem 2.5
|  $e_{1}n_{2}=n_{2}$, $e_{1}e_{1}=e_{1}$ |  |
$\mathcal{A}_{3}^{10}$ |  $n_{1}e_{1}=n_{1}$, $n_{2}e_{1}=n_{2}$ | + | Theorem 2.5
|  $e_{1}n_{1}=n_{1}$, $e_{1}n_{2}=n_{2}$, $e_{1}e_{1}=e_{1}$ |  |
$\mathcal{A}_{3}^{11}$ |  $n_{1}e_{1}=n_{2}$, $n_{2}e_{1}=n_{2}$ | + | Theorem 2.5
|  $e_{1}n_{1}=n_{2}$, $e_{1}n_{2}=n_{2}$, $e_{1}e_{1}=e_{1}$ |  |
$\mathcal{A}_{3}^{12}$ |  $n_{1}n_{1}=n_{2}$, $n_{1}e_{1}=n_{1}$, $n_{2}e_{1}=n_{2}$ | - | Theorem 2.5
|  $e_{1}n_{1}=n_{1}$, $e_{1}n_{2}=n_{2}$, $e_{1}e_{1}=e_{1}$ |  |

## References

  * [1] Ayupov Sh. A., Arzikulov F. N., Jordan counterparts of Rickart and Baer $*$-algebras, Uzbek. Mat. Zh. no. 1, 13–3 (2016)
  * [2] Ayupov Sh. A., Arzikulov F. N., Jordan counterparts of Rickart and Baer $*$-algebras, II, S̃ao Paulo J. Math. Sci. 13, 27–38 (2019)
  * [3] Arzikulov F. N., Khakimov U. I., Description of finite-dimensional inner Rickart and Baer Jordan algebras, Communications in Algebra (2023). https://doi.org/10.1080/00927872.2022.2164586
  * [4] Garces J., Li L., Peralta A., Tahlawi H., A projection-less approach to Rickart Jordan structures, Journal of Algebra 609, 567–605 (2022)
  * [5] Rakhimov I. S., Rikhsiboev I. M., Basri W., Complete lists of low dimensional complex associative algebras, arXiv:0910.0932 [math.RA] (2009). https://doi.org/10.48550/arXiv.0910.0932


On a conjecture on a series of convergence rate $\frac{1}{2}$

John M. Campbell

###### Abstract

Sun, in 2022, introduced a conjectured evaluation for a series of convergence
rate $\frac{1}{2}$ involving harmonic numbers. We prove both this conjecture
and a stronger version of this conjecture, using a summation technique based
on a beta-type integral we had previously introduced. Our full proof also
requires applications of Bailey’s
${}_{2}F_{1}\left(\frac{1}{2}\right)$-formula, Dixon’s
${}_{3}F_{2}(1)$-formula, an almost-poised version of Dixon’s formula due to
Chu, Watson’s formula for ${}_{3}F_{2}(1)$-series, the Gauss summation
theorem, Euler’s formula for ${}_{2}F_{1}$-series, elliptic integral singular
values, and lemniscate-like constants recently introduced by Campbell and Chu.
The techniques involved in our proof are useful, more broadly, in the
reduction of difficult sums of convergence rate $\frac{1}{2}$ to previously
evaluable expressions.

_MSC:_ 33C20, 33C75

_Keywords:_ harmonic number; hypergeometric series; symbolic evaluation;
closed form; digamma function

##  1 Introduction

Zhi-Wei Sun has introduced many remarkable conjectures, over the years. Many
of these conjectures are given by purported evaluations for very difficult
series that were discovered in an experimental fashion, with Computer Algebra
System software and via numerical estimates. Recently, Sun [17] posted a
preprint on series with summands involving harmonic numbers, and the purpose
of our article is to prove one of the conjectures given by Sun in this recent
preprint [17].

Conjecture 2.4 from [17] was formulated in the following manner by Sun [17],
and it was indicated [17] that this Conjecture was introduced in December of
2022. We are to let $H_{m}=1+\frac{1}{2}+\cdots+\frac{1}{m}$ denote the
$m^{\text{th}}$ entry in the sequence of harmonic numbers. The
$\Gamma$-function [16, §8] is of great importance inside of mathematics and
outside of mathematics and is to be heavily involved in our article and may be
defined via the Euler integral so that
$\Gamma(x)=\int_{0}^{\infty}u^{x-1}e^{-u}\,du$ for $\Re(x)>0$.

Conjecture 2.4 from [17]: We have

$\sum_{k=0}^{\infty}\frac{\binom{2k}{k}^{2}}{(-16)^{k}}(2H_{2k}-H_{k})=-\frac{\ln(2)\,\Gamma^{2}\left(\frac{1}{4}\right)}{4\pi\sqrt{2\pi}}$
(1)

and

$\sum_{k=0}^{\infty}\frac{\binom{2k}{k}^{2}}{32^{k}}(2H_{2k}-H_{k})=\frac{\ln(2)\,\Gamma^{2}\left(\frac{1}{4}\right)}{4\pi\sqrt{\pi}}.$
(2)

As it turns out, the alternating series evaluation in (1) can be shown to
follow in a direct way from results obtained via a hypergeometric
linearization method given by Chu and Campbell in [14], as we are to briefly
demonstrate, later in this article. However, the problem of proving the
evaluation in (2) turns out to be much more difficult. We succeed in proving
(2) in this article, as in Section 2 below. To the best of our knowledge,
Sun’s conjectured formula in (2) had not previously been proved.

###  1.1 Background

Using a linearization method based on coefficient extractions, Chu [13]
recently applied this method to obtain identities for evaluating series
involving $\frac{\binom{2k}{k}^{2}}{32^{k}}$ for $k\in\mathbb{N}_{0}$ together
with harmonic-type numbers, building on the beta integral-derived results
previously given by Campbell in [3]. However, the methds of Chu [13] cannot be
applied, at least in any direct way, to prove (2). In particular, the
hypergeometric techniques due to Chu [13] were applied in [13] to evaluate
series of convergence rate $\frac{1}{2}$ involving

$\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{k}$ (3)

for $k\in\mathbb{N}_{0}$, such as the formula

$\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}\frac{H_{k}}{k+1}=8-\frac{2\Gamma^{2}\left(\frac{1}{4}\right)}{\pi^{3/2}}-\frac{4\pi^{3/2}+16\sqrt{\pi}\ln(2)}{\Gamma^{2}\left(\frac{1}{4}\right)}$
(4)

previously introduced by Campbell [3], and series of convergence rate
$\frac{1}{2}$ involving

$\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}O_{k}^{(2)}$ (5)

for $k\in\mathbb{N}_{0}$, writing
$O_{k}^{(2)}=\sum_{i=1}^{k}\frac{1}{(2i-1)^{2}}$. However, by expanding the
summand of (2), we would need to evaluate a series of convergence rate
$\frac{1}{2}$ involving

$\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{2k}$ (6)

for $k\in\mathbb{N}_{0}$, in contrast to both (3) and (5).

There have been only a few previously known series of convergence rate
$\frac{1}{2}$ involving (6), as in our previous work on Fourier–Legendre
theory and fractional calculus [8]. In particular, it was proved in [8] that

$\displaystyle\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}\frac{H_{2k}}{2k-1}=$
(7)
$\displaystyle\frac{\sqrt{\pi}\left(\pi+3\ln(2)-4\right)}{2\Gamma^{2}\left(\frac{1}{4}\right)}-\frac{\Gamma^{2}\left(\frac{1}{4}\right)\left(\pi-3\ln(2)-2\right)}{16\pi^{3/2}}$
(8)

and that

$\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}\frac{H_{2k}}{k+1}=4-\frac{3\Gamma^{2}\left(\frac{1}{4}\right)}{2\pi^{3/2}}-\frac{2\sqrt{\pi}(\pi+3\ln(2)-4)}{\Gamma^{2}\left(\frac{1}{4}\right)},$
(9)

and the results in (7)–(8) and in (9) were highlighted as main results in [8].
Our approach toward solving the problem proposed by Sun given by proving (2)
is of a similar nature relative to our past proofs of (7)–(8) and (9), but our
proof of (2) is considerably more involved and requires recent results on what
are referred to as _lemniscate-like constants_ in [9].

As considered by Chu [13], the very fast convergence rate of series as in (4)
is of interest, especially in comparison with previously known series of
convergence rate $1$ involving $\frac{\binom{2k}{k}^{2}}{16^{k}}$ for
$k\in\mathbb{N}_{0}$ and harmonic numbers, as in the formula

$\sum_{k=0}^{\infty}\left(\frac{1}{16}\right)^{k}\binom{2k}{k}^{2}\frac{H_{k}}{(2k-1)^{2}}=\frac{12-16\ln(2)}{\pi}$

proved by Choi in 2014 [11] and independently by Chen in 2016 [10], which was
later generalized by Campbell [4] and by Wang and Chu [20]. The foregoing
considerations greatly motivate the development of new techniques for
evaluating series involving expressions as in (3).

###  1.2 Preliminaries

One of the key tools that we are to apply to prove Sun’s conjectured formula
shown in (2) is given by the beta-type integration technique introduced in the
author’s previous publication [5] and reproduced in the author’s PhD Thesis
[6]. We are to let $H_{m}^{\prime}=1-\frac{1}{2}+\cdots+\frac{(-1)^{m+1}}{m}$
denote the $m^{\text{th}}$ entry in the sequence of alternating harmonic
numbers, recalling the relation such that $H_{2n}^{\prime}=H_{2n}-H_{n}$.

######  Lemma 1.

Given a sequence $(f_{n})_{n\geq 0}$ whereby the series

$\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}H_{2n}^{\prime}\binom{2n}{n}^{2}\frac{f_{n}}{n+1}$
(10)

converges, the above series is equal to

$\displaystyle\frac{4}{\pi}\int_{0}^{1}\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\sqrt{1-x^{2}}\binom{-\frac{1}{2}}{n}f_{n}\ln\left(x\right)\,dx$
(11)
$\displaystyle+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{16}\right)^{n}\frac{\binom{2n}{n}^{2}(2\ln(2)(n+1)+1)}{(n+1)^{2}}f_{n},$
(12)

under the assumption that the sequence $f$ is such that it is possible to
reverse the order of integration and infinite summation in (11) [5, 6].

The evaluation of hypergeometric series is of central importance in our
article, so we find it to be appropriate to recall the following definition of
the term _generalized hypergeometric series_ :

${}_{p}F_{q}\\!\\!\left[\begin{matrix}a_{1},a_{2},\ldots,a_{p}\vspace{1mm}\\\
b_{1},b_{2},\ldots,b_{q}\end{matrix}\ \Bigg{|}\
x\right]=\sum_{n=0}^{\infty}\left[\begin{matrix}a_{1},a_{2},\ldots,a_{p}\vspace{1mm}\\\
b_{1},b_{2},\ldots,b_{q}\end{matrix}\right]_{n}\frac{x^{n}}{n!}.$ (13)

The _complete elliptic integrals_ are to play a significant role in our proof
of (2). For the purposes of our article, the complete elliptic integrals K and
E of the first and second kinds may be defined via the following Maclaurin
series expansions, with reference to the classic _Pi and the AGM_ text [2, pp.
8–10]:

$\displaystyle\text{{\bf
K}}(k)=\frac{\pi}{2}\cdot{}_{2}F_{1}\\!\\!\left[\begin{matrix}\frac{1}{2},\frac{1}{2}\vspace{1mm}\\\
1\end{matrix}\ \Bigg{|}\ k^{2}\right],$ (14) $\displaystyle\text{{\bf
E}}(k)=\frac{\pi}{2}\cdot{}_{2}F_{1}\\!\\!\left[\begin{matrix}\frac{1}{2},-\frac{1}{2}\vspace{1mm}\\\
1\end{matrix}\ \Bigg{|}\ k^{2}\right].$ (15)

The following elliptic integral singular values highlighted as Theorem 1.7 in
the _Pi and the AGM_ text [2, p. 25] are to also be applied in our proof of
(2):

$\displaystyle\text{{\bf
K}}\left(\frac{1}{\sqrt{2}}\right)=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{4\sqrt{\pi}},$
(16) $\displaystyle\text{{\bf
E}}\left(\frac{1}{\sqrt{2}}\right)=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{8\sqrt{\pi}}+\frac{\pi^{3/2}}{\Gamma^{2}\left(\frac{1}{4}\right)}.$
(17)

Writing $k^{\prime}=\sqrt{1-k^{2}}$, we are to also employ the following
differential equations [2, p. 10]:

$\frac{d\text{{\bf E}}}{dk}=\frac{\text{{\bf E}}-\text{{\bf K}}}{k}\ \ \
\text{and}\ \ \ \frac{d\text{{\bf K}}}{dk}=\frac{\text{{\bf
E}}-(k^{\prime})^{2}\text{{\bf K}}}{k(k^{\prime})^{2}}.$ (18)

We are to make use of the _Gauss’s summation theorem_ [1, §1.3] such that

${}_{2}F_{1}\\!\\!\left[\begin{matrix}a,b\vspace{1mm}\\\ c\end{matrix}\
\Bigg{|}\ 1\right]=\Gamma\left[\begin{matrix}c,c-a-b\vspace{1mm}\\\
c-a,c-b\end{matrix}\right]$ (19)

for $\Re(c-a-b)>0$, writing

$\Gamma\left[\begin{matrix}\alpha,\beta,\ldots,\gamma\vspace{1mm}\\\
A,B,\ldots,C\end{matrix}\right]=\frac{\Gamma(\alpha)\Gamma(\beta)\cdots\Gamma(\gamma)}{\Gamma(A)\Gamma(B)\cdots\Gamma(C)}.$

The _Euler–Mascheroni constant_ is such that
$\gamma=\lim_{n\to\infty}\big{(}H_{n}-\ln n\big{)}$. The _digamma function_ is
the special function such that [16, §9]:

$\psi(z)=\frac{d}{dz}\ln\Gamma(z)=\frac{\Gamma^{\prime}(z)}{\Gamma(z)}=-\gamma+\sum_{n=0}^{\infty}\frac{z-1}{(n+1)(n+z)}.$
(20)

Euler’s formula for ${}_{2}F_{1}$-series [15, p. 57] is such that

$\displaystyle{}_{2}F_{1}\\!\\!\left[\begin{matrix}a,b\vspace{1mm}\\\
c\end{matrix}\ \Bigg{|}\ z\right]=\Gamma\left[\begin{matrix}c\vspace{1mm}\\\
c-a,a\end{matrix}\right]\int_{0}^{1}t^{a-1}(1-t)^{c-a-1}(1-zt)^{-b}\,dt,$

for $\Re(c)>\Re(a)>0$ and for $|\text{arg}(1-z)|<\pi$. Following [15, p. 111],
Euler’s formula may be applied to obtain hypergeometric reduction formula

$z{}_{3}F_{2}\\!\\!\left[\begin{matrix}a,b,1\vspace{1mm}\\\ c,2\end{matrix}\
\Bigg{|}\
z\right]=\frac{c-1}{(a-1)(b-1)}\left({}_{2}F_{1}\\!\\!\left[\begin{matrix}a-1,b-1\vspace{1mm}\\\
c-1\end{matrix}\ \Bigg{|}\ z\right]-1\right)$ (21)

given in [15, p. 111]. So, the Gauss identity in (19) gives us, from (21),
that

${}_{3}F_{2}\\!\\!\left[\begin{matrix}a,b,1\vspace{1mm}\\\ c,2\end{matrix}\
\Bigg{|}\
1\right]=\frac{c-1}{(a-1)(b-1)}\left(\Gamma\left[\begin{matrix}c-1,c-a-b+1\vspace{1mm}\\\
c-a,c-b\end{matrix}\right]-1\right)$ (22)

for $a\neq 1$, $b\neq 1$, and $\Re(c-a-b)>-1$ [15, p. 111]. By L’Hôpital’s
rule, we may obtain from (22) that

${}_{3}F_{2}\\!\\!\left[\begin{matrix}a,1,1\vspace{1mm}\\\ c,2\end{matrix}\
\Bigg{|}\ 1\right]=\frac{c-1}{a-1}\left(\psi(c-1)-\psi(c-a)\right)$ (23)

for $a\neq 1$ and $\Re(c-a)>0$ [15, p. 111]. The identity in (23) is to be
required in a direct way in our proof of (2).

The evaluation

${}_{3}F_{2}\\!\\!\left[\begin{matrix}\frac{1}{2},1,\frac{5}{4}\vspace{1mm}\\\
\frac{3}{2},\frac{7}{4}\end{matrix}\ \Bigg{|}\
1\right]=3-\frac{6\pi^{3}}{\Gamma^{4}\left(\frac{1}{4}\right)}$ (24)

was introduced by Campbell et al. in [8] and highlighted as part of a main
Theorem from [8]. This evaluation in (24) is to provide a key tool in our
proof of Sun’s conjectured formula in (2). The ${}_{3}F_{2}(1)$-evaluation in
(24) was proved in [8] using FL theory and fractional operators. The
evaluation in (24) was applied in a similar context in [9], in which (24) was
proved using a reindexing argument together with Watson’s formula for
${}_{3}F_{2}(1)$-series [1, §1.3]:

${}_{3}F_{2}\\!\\!\left[\begin{matrix}a,b,c\vspace{1mm}\\\
\frac{a+b+1}{2},2c\end{matrix}\ \Bigg{|}\
1\right]=\Gamma\left[\begin{matrix}\frac{1}{2},c+\frac{1}{2},\frac{a+b+1}{2},c-\frac{a+b-1}{2}\vspace{1mm}\\\
\frac{a+1}{2},\frac{b+1}{2},c-\frac{a-1}{2},c-\frac{b-1}{2}\end{matrix}\right].$
(25)

Indeed, by following [9], we may apply a reindexing argument to the series in
(24) so that

$\displaystyle{}_{3}F_{2}\\!\\!\left[\begin{matrix}\frac{1}{2},1,\frac{5}{4}\vspace{1mm}\\\
\frac{3}{2},\frac{7}{4}\end{matrix}\ \Bigg{|}\ 1\right]$
$\displaystyle=\sum_{k=0}^{\infty}\frac{\left(\frac{1}{2}\right)_{k}\left(\frac{5}{4}\right)_{k}}{\left(\frac{3}{2}\right)_{k}\left(\frac{7}{4}\right)_{k}}$
$\displaystyle=-3\sum_{k=1}^{\infty}\frac{\left(-\frac{1}{2}\right)_{k}\left(\frac{1}{4}\right)_{k}}{\left(\frac{1}{2}\right)_{k}\left(\frac{3}{4}\right)_{k}}$
$\displaystyle=3\left(1-{}_{3}F_{2}\\!\\!\left[\begin{matrix}1,-\frac{1}{2},\frac{1}{4}\vspace{1mm}\\\
\frac{1}{2},\frac{3}{4}\end{matrix}\ \Bigg{|}\ 1\right]\right),$

and so that the formulation of Watson’s identity in (25) then gives us the
desired symbolic form in (24).

We let

$\beta(x,y)=\int_{0}^{1}t^{x-1}(1-t)^{y-1}\,dt$

denote the beta function for $\Re(x)>0$ and $\Re(y)>0$. The classical
_lemniscate constants_

$A=\frac{1}{4}\beta\left(\frac{1}{2},\frac{1}{4}\right)=\int_{0}^{1}\frac{1}{\sqrt{1-t^{4}}}\,dt=\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+1}=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{4\sqrt{2\pi}}$
(26)

and

$B=\frac{1}{4}\beta\left(\frac{1}{2},\frac{3}{4}\right)=\int_{0}^{1}\frac{t^{2}}{\sqrt{1-t^{4}}}\,dt=\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+3}=\frac{\sqrt{2\pi^{3}}}{\Gamma^{2}\left(\frac{1}{4}\right)}$
(27)

have been of much significance in the history of mathematics [9, 19], and this
led to the exploration of _lemniscate-like constants_ of the following forms
[7, 9], for a suitable sequence $f$:

$\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{f_{n}}{4n+1}\
\ \ \text{and}\ \ \
\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{f_{n}}{4n+3}.$
(28)

Series as in (28) are to be heavily used in our main proof in Section 2.

_Dixon’s formula_ for well-poised series is such that

${}_{3}F_{2}\\!\\!\left[\begin{matrix}a,b,c\vspace{1mm}\\\
1+a-b,1+a-c\end{matrix}\ \Bigg{|}\
1\right]=\Gamma\left[\begin{matrix}1+\frac{a}{2},1+\frac{a}{2}-b-c,1+a-b,1+a-c\vspace{1mm}\\\
1+a,1+a-b-c,1+\frac{a}{2}-b,1+\frac{a}{2}-c\end{matrix}\right].$ (29)

Chu [12] introduced extended versions of Watson–Whipple–Dixon
${}_{3}F_{2}$-series, and we are to apply the following almost-poised version
of Dixon’s formula [12] as in [9]:

$\displaystyle{}_{3}F_{2}\\!\\!\left[\begin{matrix}a,b,c\vspace{1mm}\\\
2+a-b,2+a-c\end{matrix}\ \Bigg{|}\ 1\right]=$
$\displaystyle\frac{2^{1+2a-2b-2c}\Gamma(a-b+2)\Gamma(a-c+2)}{\pi(b-1)(1-c)\Gamma(a)\Gamma(a-2b+2)\Gamma(a-2c+2)\Gamma(a-b-c+2)}$
$\displaystyle\Bigg{(}\Gamma\left(\frac{1+a}{2}\right)\Gamma\left(\frac{2+a}{2}-b\right)\Gamma\left(\frac{2+a}{2}-c\right)\Gamma\left(\frac{5+a}{2}-b-c\right)$
$\displaystyle-\Gamma\left(\frac{a}{2}\right)\Gamma\left(\frac{3+a}{2}-b\right)\Gamma\left(\frac{3+a}{2}-c\right)\Gamma\left(\frac{4+a}{2}-b-c\right)\Bigg{)}.$

A direct application of the above hypergeometric identity gives us that:

$\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+3)^{2}}=\frac{1}{9}\,{}_{3}F_{2}\\!\\!\left[\begin{matrix}\frac{1}{2},\frac{3}{4},\frac{3}{4}\vspace{1mm}\\\
\frac{7}{4},\frac{7}{4}\end{matrix}\ \Bigg{|}\
1\right]=\frac{4-\pi}{4\sqrt{2\pi}}\Gamma^{2}\left(\frac{3}{4}\right).$ (30)

The lemniscate-like constant evaluation such that

$\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+3}=\frac{\pi^{3/2}(3\ln(2)+2)}{2\sqrt{2}\Gamma^{2}\left(\frac{1}{4}\right)}$
(31)

and

$\sum_{k=0}^{\infty}\left(\frac{1}{4}\right)^{k}\binom{2k}{k}\frac{O_{2k}}{4k+1}=\frac{3\Gamma^{2}\left(\frac{1}{4}\right)\ln(2)}{16\sqrt{2\pi}}$
(32)

were proved in [9] and included as main results, and we are to apply (31) and
(32) in our main proof, letting $O_{m}=1+\frac{1}{3}+\cdots+\frac{1}{2m-1}$
denote the $m^{\text{th}}$ odd harmonic number.

_Bailey’s theorem_ [1, p. 11] is such that

${}_{2}F_{1}\\!\\!\left[\begin{matrix}a,1-a\vspace{1mm}\\\ c\end{matrix}\
\Bigg{|}\
\frac{1}{2}\right]=\Gamma\left[\begin{matrix}\frac{c}{2},\frac{c+1}{2}\vspace{1mm}\\\
\frac{a+c}{2},\frac{1-a+c}{2}\end{matrix}\right].$ (33)

Tauraso [18] obtained the following, by setting $a=\frac{1}{2}$ in (33) and
via a term-by-term application of the operator $\frac{\partial}{\partial
c}\cdot\big{|}_{c=1}$:

$\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{k}=\frac{\sqrt{\pi}(\pi-4\ln(2))}{2\Gamma^{2}\left(\frac{3}{4}\right)}.$
(34)

However, the application of operators such as $\frac{\partial}{\partial
c}\cdot\big{|}_{c=1/2}$ have the effect of reducing the power of central
binomial coefficients, so it is unclear as to how it may be possible to mimic
Tauraso’s approach in the hope of evaluating the following intractable series:

$\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{2k}.$

##  2 Proof of a conjectured evaluation for a series of convergence rate
$\frac{1}{2}$

######  Theorem 1.

The formula in (2) conjectured by Sun holds true.

###### Proof.

We begin by setting $f_{n}=2^{-n}(n+1)$ in Lemma 1. Lemma 1 then gives us the
equality of

$\sum_{n=0}^{\infty}\left(\frac{1}{32}\right)^{n}\left(H_{2n}-H_{n}\right)\binom{2n}{n}^{2}$
(35)

and

$\displaystyle\frac{4}{\pi}\int_{0}^{1}\sqrt{1-x^{2}}\ln(x)\sum_{n=0}^{\infty}\left(-\frac{x^{2}}{2}\right)^{n}\binom{-\frac{1}{2}}{n}(n+1)\,dx+$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{32}\right)^{n}\binom{2n}{n}^{2}\frac{1+2(n+1)\ln(2)}{n+1}.$

According to the Maclaurin series expansions in (14) and (15) along with the
differential equations in (18), we may obtain the power series expansion

$\sum_{n=0}^{\infty}\frac{\binom{2n}{n}^{2}}{n+1}y^{n}=\frac{\text{{\bf
E}}\left(4\sqrt{y}\right)}{4\pi
y}+\frac{4\left(1-\frac{1}{16y}\right)\text{{\bf
K}}\left(4\sqrt{y}\right)}{\pi}.$ (36)

From the expansion in (36) together with the elliptic integral singular values
shown in (16) and in (17), we find that the series in (35) is expressible in
the following manner:

$\displaystyle\frac{4}{\pi}\int_{0}^{1}\sqrt{1-x^{2}}\ln(x)\sum_{n=0}^{\infty}\left(-\frac{x^{2}}{2}\right)^{n}\binom{-\frac{1}{2}}{n}(n+1)\,dx+$
(37)
$\displaystyle\frac{4\sqrt{\pi}}{\Gamma^{2}\left(\frac{1}{4}\right)}+\frac{\ln(2)\Gamma^{2}\left(\frac{1}{4}\right)}{2\pi^{3/2}}.$
(38)

By the generalized binomial theorem, we find that (37)–(38) is reducible to
the following:

$-\frac{4}{\pi\sqrt{2}}\int_{0}^{1}\frac{\sqrt{1-x^{2}}\left(x^{2}-4\right)\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx+\frac{4\sqrt{\pi}}{\Gamma^{2}\left(\frac{1}{4}\right)}+\frac{\Gamma^{2}\left(\frac{1}{4}\right)\ln(2)}{2\pi^{3/2}}.$
(39)

So, it remains to evaluate the integral in (39), i.e., to evaluate the
following expression:

$\int_{0}^{1}\frac{\sqrt{1-x^{2}}\left(x^{2}-4\right)\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx.$
(40)

We may rewrite (40) as

$-\int_{0}^{1}\sqrt{\frac{1-x^{2}}{2-x^{2}}}\ln(x)\,dx-2\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx.$
(41)

Applying the change of variables such that $1-x^{2}=u$ to the first integral
in (41), we obtain

$-\frac{1}{4}\int_{0}^{1}\frac{\sqrt{u}\ln(1-u)}{\sqrt{1-u^{2}}}\,du-2\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx.$
(42)

By expanding the integrand factor $\ln(1-u)$ with its Maclaurin series and
then integrating term-by-term using the Dominated Convergence Theorem, we may
obtain the following from (42):

$\frac{\sqrt{\pi}}{8}\sum_{n=1}^{\infty}\frac{1}{n}\Gamma\left[\begin{matrix}\frac{n}{2}+\frac{3}{4}\vspace{1mm}\\\
\frac{n}{2}+\frac{5}{4}\end{matrix}\right]-2\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx.$
(43)

Applying a series bisection to (43), we obtain

$\displaystyle\frac{\sqrt{\pi}}{16}\sum_{n=1}^{\infty}\frac{1}{n}\Gamma\left[\begin{matrix}n+\frac{3}{4}\vspace{1mm}\\\
n+\frac{5}{4}\end{matrix}\right]+\frac{\sqrt{\pi}}{8}\sum_{n=1}^{\infty}\frac{1}{2n-1}\Gamma\left[\begin{matrix}n+\frac{1}{4}\vspace{1mm}\\\
n+\frac{3}{4}\end{matrix}\right]-$ (44) $\displaystyle
2\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx.$

Applying an index shift to the first series in (44), we obtain:

$\sum_{n=1}^{\infty}\frac{1}{n}\Gamma\left[\begin{matrix}n+\frac{3}{4}\vspace{1mm}\\\
n+\frac{5}{4}\end{matrix}\right]=\frac{12\Gamma\left(\frac{3}{4}\right)}{5\Gamma\left(\frac{1}{4}\right)}{}_{3}F_{2}\\!\\!\left[\begin{matrix}1,1,\frac{7}{4}\vspace{1mm}\\\
2,\frac{9}{4}\end{matrix}\ \Bigg{|}\ 1\right].$ (45)

So, the digamma identity in (23) from [15, p. 111] allows us to evaluate the
${}_{3}F_{2}$-expression in (45). This allows us to rewrite (42) in the
following manner:

$\displaystyle-\frac{\pi^{3/2}(\pi+2\ln(2)-8)}{4\sqrt{2}\Gamma^{2}\left(\frac{1}{4}\right)}+\frac{\sqrt{\pi}}{8}\sum_{n=1}^{\infty}\frac{1}{2n-1}\Gamma\left[\begin{matrix}n+\frac{1}{4}\vspace{1mm}\\\
n+\frac{3}{4}\end{matrix}\right]-$ (46) $\displaystyle
2\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx.$
(47)

By rewriting the infinite series in (46) according to the notation in (13), we
find that (46)–(47) is equal to:

$\displaystyle-\frac{\pi^{3/2}(\pi+2\ln(2)-8)}{4\sqrt{2}\Gamma^{2}\left(\frac{1}{4}\right)}+\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{24\sqrt{2\pi}}{}_{3}F_{2}\\!\\!\left[\begin{matrix}\frac{1}{2},1,\frac{5}{4}\vspace{1mm}\\\
\frac{3}{2},\frac{7}{4}\end{matrix}\ \Bigg{|}\ 1\right]-$ $\displaystyle
2\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx.$

So, from the Watson-derived ${}_{3}F_{2}(1)$-evaluation on display in (24), we
find that the integral in (40) is equal to the following:

$\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{8\sqrt{2\pi}}-\frac{\pi^{3/2}(\pi+\ln(2)-4)}{2\sqrt{2}\Gamma^{2}\left(\frac{1}{4}\right)}-2\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx.$
(48)

So, it remains to evaluate the integral in (48). Using a change of variables,
we obtain that:

$\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=\frac{1}{4}\int_{0}^{1}\frac{\sqrt{u}\ln(1-u)}{(1+u)^{3/2}\sqrt{1-u}}\,du.$

Using an appropriate Cauchy product, we may obtain that

$\left(\left(\frac{d}{du}\right)^{n}\frac{1}{(u+1)\sqrt{1-u^{2}}}\right)\,\Bigg{|}_{u=0}=\left(-\frac{1}{2}\right)^{n}(n+1)!\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}.$
(49)

So, from the Maclaurin series corresponding to (49), we may obtain that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
(50)
$\displaystyle\frac{1}{4}\int_{0}^{1}\left(\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}(n+1)u^{n+\frac{1}{2}}\ln(1-u)\right)\,du.$
(51)

According to a standard moment formula for the digamma function, we have that:

$\int_{0}^{1}u^{n+\frac{1}{2}}\ln(1-u)\,du=-\frac{2\left(\psi\left(n+\frac{5}{2}\right)+\gamma\right)}{2n+3}.$
(52)

According to the expansion formula for the digamma function shown in (20), we
may obtain from (52) that

$\int_{0}^{1}u^{n+\frac{1}{2}}\ln(1-u)\,du=\frac{4\ln(2)-4O_{n+1}-\frac{4}{2n+3}}{2n+3}.$
(53)

According to the Dominated Convergence Theorem, we may reverse the order of
integration and infinite summation with respect to the equality in (50)–(51),
so as to give us the following, being consistent with the notation in (53):

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{1}{4}\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}\frac{(n+1)\left(4\ln(2)-4O_{n+1}-\frac{4}{2n+3}\right)}{2n+3}.$

Applying a series bisection, we obtain that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{(2n+1)\left(4\ln(2)-4O_{2n+1}-\frac{4}{2(2n)+3}\right)}{2(2n)+3}-$
(54)
$\displaystyle\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{2n+1}\binom{2n+1}{n}\frac{((2n+1)+1)\left(4\ln(2)-4O_{(2n+1)+1}-\frac{4}{2(2n+1)+3}\right)}{2(2n+1)+3}.$

Equivalently,

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle-\frac{1}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+1}+\left(-1-\frac{\ln(2)}{2}\right)\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+3}+$
$\displaystyle\frac{13+12\ln(2)}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+5}+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+3)^{2}}-$
$\displaystyle\frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+5)^{2}}+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+3}-$
$\displaystyle\frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+5}.$

From (26) and (27), we may obtain that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle-\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{32\sqrt{2\pi}}+\frac{\left(-1-\frac{\ln(2)}{2}\right)\sqrt{2\pi^{3}}}{\Gamma^{2}\left(\frac{1}{4}\right)}+$
$\displaystyle\frac{13+12\ln(2)}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+5}+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+3)^{2}}-$
$\displaystyle\frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+5)^{2}}+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+3}-$
$\displaystyle\frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+5}.$

From the lemniscate-like constant evaluations in (30) and (31), we obtain that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle-\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{32\sqrt{2\pi}}+\frac{(4-\pi)\Gamma^{2}\left(\frac{3}{4}\right)}{8\sqrt{2\pi}}+\frac{\pi^{3/2}(3\ln(2)+2)}{4\sqrt{2}\Gamma^{2}\left(\frac{1}{4}\right)}+\frac{\left(-1-\frac{\ln(2)}{2}\right)\sqrt{2\pi^{3}}}{\Gamma^{2}\left(\frac{1}{4}\right)}+$
$\displaystyle\frac{13+12\ln(2)}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+5}-\frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+5)^{2}}-$
$\displaystyle\frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+5}.$

Applying reindexing arguments, we obtain that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{-\Gamma^{4}\left(\frac{1}{4}\right)-8\pi^{2}(2+\pi+\ln(2))}{32\sqrt{2\pi}\Gamma\left(\frac{1}{4}\right)^{2}}+$
$\displaystyle\frac{3+4\ln(2)}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{2n-1}+\left(\frac{1}{2}+\frac{\ln(2)}{2}\right)\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+1}-$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+1)^{2}}+\frac{9}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-3}-$
$\displaystyle\frac{3}{4}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n-1)}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{2n-1}-$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+1}.$

From an equivalent formulation of the generating function for the sequence of
Catalan numbers, we obtain that:

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{-\Gamma^{4}\left(\frac{1}{4}\right)-8\pi^{2}(2+\pi+\ln(2))}{32\sqrt{2\pi}\Gamma^{2}\left(\frac{1}{4}\right)}+$
$\displaystyle\left(\frac{1}{2}+\frac{\ln(2)}{2}\right)\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+1}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+1)^{2}}+$
$\displaystyle\frac{9}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-3}+\sum_{n=0}^{\infty}-\frac{3}{4}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-1}-$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{2n-1}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+1}.$

From the classical lemniscate constant evaluation shown in (26), we obtain
that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{-\Gamma^{4}\left(\frac{1}{4}\right)-8\pi^{2}(2+\pi+\ln(2))}{32\sqrt{2\pi}\Gamma^{2}\left(\frac{1}{4}\right)}+\frac{\left(\frac{1}{2}+\frac{\ln(2)}{2}\right)\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}-$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n+1)^{2}}+\frac{9}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-3}-$
$\displaystyle\frac{3}{4}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{(4n-1)}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{2n-1}-$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+1}.$

From Dixon’s formula in (29), we obtain that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{2\pi\Gamma\left(-\frac{1}{4}\right)(2+\pi+\ln(2))+\sqrt{2}\Gamma\left(\frac{1}{4}\right)^{3}(3-\pi+4\ln(2))}{64\sqrt{\pi}\Gamma\left(\frac{1}{4}\right)}+$
$\displaystyle\frac{9}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-3}-\frac{3}{4}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-1}-$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{2n-1}-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{4n+1}.$

From the lemniscate-like constant evaluation in (32), we obtain that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{\sqrt{2}\Gamma\left(\frac{1}{4}\right)^{3}(3-\pi+\ln(2))+2\pi\Gamma\left(-\frac{1}{4}\right)(2+\pi+\ln(2))}{64\sqrt{\pi}\Gamma\left(\frac{1}{4}\right)}+$
$\displaystyle\frac{9}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-3}-\frac{3}{4}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-1}-$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{O_{2n}}{2n-1}.$

Using the moment formula

$O_{2n}=\int_{0}^{1}\frac{1-x^{4n}}{1-x^{2}}\,dx,$

together with the power series identity

$\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1-x^{4n}}{(2n-1)\left(1-x^{2}\right)}=-\frac{\sqrt{1-x^{4}}}{x^{2}-1},$

we may oobtain that

$\int_{0}^{1}-\frac{\sqrt{1-x^{4}}}{x^{2}-1}\,dx=\text{{\bf E}}(i),$

so that we may apply a known elliptic integral singular value to evaluate the
remaining harmonic sum. Explicitly,

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{-48\pi^{2}-8\pi^{3}-8\pi^{2}\ln(2)+\Gamma^{4}\left(\frac{1}{4}\right)(-1-\pi+\ln(2))}{32\sqrt{2\pi}\Gamma^{2}\left(\frac{1}{4}\right)}+$
$\displaystyle\frac{9}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-3}-\frac{3}{4}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n-1}.$

Applying reindexing arguments, we obtain that

$\displaystyle\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=$
$\displaystyle\frac{3}{8}-\frac{\pi^{3/2}(6+\pi+\ln(2))}{4\sqrt{2}\Gamma^{2}\left(\frac{1}{4}\right)}+\frac{\Gamma^{2}\left(\frac{1}{4}\right)(-1-\pi+\ln(2))}{32\sqrt{2\pi}}-$
$\displaystyle\frac{3}{16}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{n+1}+\frac{3}{8}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+1}+$
$\displaystyle\frac{3}{4}\sum_{n=0}^{\infty}\left(\frac{1}{4}\right)^{n}\binom{2n}{n}\frac{1}{4n+3}$

Using the generating function for the sequence of Catalan numbers, together
with the evaluations for the classical lemniscate constants, we obtain

$\int_{0}^{1}\frac{\sqrt{1-x^{2}}\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)(4-2\pi+2\ln(2))}{64\sqrt{2\pi}}-\frac{\pi^{3/2}(\pi+\ln(2))}{4\sqrt{2}\Gamma^{2}\left(\frac{1}{4}\right)}.$

So, from (48), we find that the integral in (40) may be reduced to the
following:

$\int_{0}^{1}\frac{\sqrt{1-x^{2}}\left(x^{2}-4\right)\ln(x)}{\left(2-x^{2}\right)^{3/2}}\,dx=\frac{\sqrt{2}\pi^{3/2}}{\Gamma^{2}\left(\frac{1}{4}\right)}+\left(\frac{\sqrt{\frac{\pi}{2}}}{16}-\frac{\ln(2)}{16\sqrt{2\pi}}\right)\Gamma^{2}\left(\frac{1}{4}\right).$

So, from the equality of (35) and (39), we obtain the equality

$\sum_{n=0}^{\infty}\left(\frac{1}{32}\right)^{n}\left(H_{2n}-H_{n}\right)\binom{2n}{n}^{2}=\frac{(5\ln(2)-\pi)\Gamma^{2}\left(\frac{1}{4}\right)}{8\pi^{3/2}}.$

So, from the harmonic sum evaluation in (34) derived from Bailey’s theorem, we
obtain that:

$\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{2k}=\frac{(\pi-3\ln(2))\Gamma^{2}\left(\frac{1}{4}\right)}{8\pi^{3/2}}.$
(55)

So, by taking an appropriate linear combination of (55) and the formula in
(34) derived from Bailey’s theorem, we obtain the desired result. ∎

The formulas

$\sum_{k=0}^{\infty}\left(-\frac{1}{16}\right)^{k}\binom{2k}{k}^{2}H_{k}=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{4\sqrt{2\pi^{3}}}\left(\pi-5\ln(2)\right)$

and

$\sum_{k=0}^{\infty}\left(-\frac{1}{16}\right)^{k}\binom{2k}{k}^{2}H_{2k}=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{8\sqrt{2\pi^{3}}}\left(\pi-6\ln(2)\right)$

were highlighted as part of Theorem 4 in [14] and proved via a linearization
method introduced in [14]. So, by taking an appropriate linear combination of
the above formulas, this gives us a proof of Sun’s conjectured formula in (1).

The new formula in (55) is of interest in its own right and recalls the open
problem recently considered in [13] to evaluate

$\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{k}^{(2)}\
\ \ \text{and}\ \ \
\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{k}^{2}$

symbolically, writing $H_{m}^{(2)}=1+\frac{1}{2^{2}}+\cdots+\frac{1}{m^{2}}$.
Our evaluation in (55) also motivates our interest in the open problem of
evaluating

$\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{2k}^{(2)}\
\ \ \text{and}\ \ \
\sum_{k=0}^{\infty}\left(\frac{1}{32}\right)^{k}\binom{2k}{k}^{2}H_{2k}^{2}$

symbolically.

## References

  * [1] W. N. Bailey, Generalized Hypergeometric Series, Cambridge University Press, Cambridge (1935).
  * [2] J. M. Borwein and P. B. Borwein, Pi and the AGM, John Wiley & Sons, Inc., New York (1987).
  * [3] J. M. Campbell, New series involving harmonic numbers and squared central binomial coefficients, _Rocky Mt. J. Math._ 49(8) (2019), 2513–2544.
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John M. Campbell

Department of Mathematics

Toronto Metropolitan University

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