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# Mneimneh-type Binomial Sums of Multiple Harmonic-type Sums

Ende Pana, and Ce Xub,
a. College of Teacher Education, Quzhou University,
Quzhou 324022, P.R. China
b. School of Mathematics and Statistics, Anhui Normal University,
Wuhu 241002, P.R. China  Email<EMAIL_ADDRESS><EMAIL_ADDRESS>
Abstract. In this paper, we establish some expressions of Mneimneh-type
binomial sums involving multiple harmonic-type sums in terms of finite sums of
Stirling numbers, Bell numbers and some related variables. In particular, we
present some new formulas of Mneimneh-type binomial sums involving generalized
(alternating) harmonic numbers. Further, we establish a new identity relating
the multiple zeta star values $\zeta^{\star}(m+2,\\{1\\}_{r-1})$ and specific
multiple polylogarithms by applying the Toeplitz principle. Furthermore, we
present some interesting consequences and illustrative examples.

Keywords: Mneimneh’s identity; harmonic numbers; multiple zeta (star) values;
multiple polylogarithms; multiple harmonic (star) sums; Binomial coefficients.

AMS Subject Classifications (2020): 11M32, 11M99.

##  1 Introduction

We begin with some basic notations. Let $\mathbb{N}$ be the set of positive
integers and $\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}$. A finite sequence
${\boldsymbol{\sl{k}}}:=(k_{1},\ldots,k_{r})\in\mathbb{N}^{r}$ is called a
_composition_. We put

$|{\boldsymbol{\sl{k}}}|:=k_{1}+\cdots+k_{r},\quad\operatorname*{dep}({\boldsymbol{\sl{k}}}):=r,$

and call them the weight and the depth of ${\boldsymbol{\sl{k}}}$,
respectively. If $k_{1}>1$, ${\boldsymbol{\sl{k}}}$ is called _admissible_. As
a convention, we denote by $\\{1\\}_{d}$ the sequence of 1’s with $d$
repetitions.

For a composition ${\boldsymbol{\sl{k}}}=(k_{1},\ldots,k_{r})$ and positive
integer $n$, the _multiple harmonic sums_ (MHSs) and _multiple harmonic star
sums_ (MHSSs) are defined by

$\displaystyle\zeta_{n}({\boldsymbol{\sl{k}}}):=\displaystyle\sum\limits_{n\geq
n_{1}>\cdots>n_{r}>0}\displaystyle\frac{1}{n_{1}^{k_{1}}\cdots
n_{r}^{k_{r}}}\quad\text{and}\quad\zeta^{\star}_{n}({\boldsymbol{\sl{k}}}):=\displaystyle\sum\limits_{n\geq
n_{1}\geq\cdots\geq n_{r}>0}\displaystyle\frac{1}{n_{1}^{k_{1}}\cdots
n_{r}^{k_{r}}},$ (1.1)

respectively. If $n<r$ then ${\zeta_{n}}({\boldsymbol{\sl{k}}}):=0$ and
${\zeta_{n}}(\emptyset)={\zeta^{\star}_{n}}(\emptyset):=1$. In particular, if
${\boldsymbol{\sl{k}}}=(k)\in\mathbb{N}$ in (1.1), then

$\displaystyle\zeta_{n}(k)=\zeta^{\star}_{n}(k)\equiv
H_{n}^{(k)}=\displaystyle\sum_{j=1}^{n}\displaystyle\frac{1}{j^{k}}$ (1.2)

is the $n$-th generalized _harmonic number_ of order $k$, and furthermore, if
$k=1$ then $H_{n}\equiv H_{n}^{(1)}$ is the classical $n$-th harmonic number.
When taking the limit $n\rightarrow\infty$ in (1.1) we get the so-called the
_multiple zeta values_ (MZVs) and the _multiple zeta star values_ (MZSVs),
respectively ([5, 18, 19])

$\displaystyle{\zeta}({\boldsymbol{\sl{k}}}):=\displaystyle\lim_{n\rightarrow\infty}{\zeta_{n}}({\boldsymbol{\sl{k}}})\quad\text{and}\quad{\zeta^{\star}}({\boldsymbol{\sl{k}}}):=\displaystyle\lim_{n\rightarrow\infty}{\zeta^{\star}_{n}}({\boldsymbol{\sl{k}}}),$

defined for an admissible composition ${\boldsymbol{\sl{k}}}$ to ensure
convergence of the series. More generally, let
${\boldsymbol{\sl{k}}}=(k_{1},\ldots,k_{r})\in\mathbb{N}^{r}$ and
${\boldsymbol{\sl{x}}}=(x_{1},\dotsc,x_{r})$ where $x_{1},\dotsc,x_{r}$ are
complex variables. The classical _multiple polylogarithm_ (MPL) and _multiple
polylogarithm star function_ with $r$ variables are defined by

$\displaystyle\operatorname{Li}_{{\boldsymbol{\sl{k}}}}({\boldsymbol{\sl{x}}}):=\displaystyle\sum_{n_{1}>n_{2}>\cdots>n_{r}>0}\displaystyle\frac{x_{1}^{n_{1}}\dotsm
x_{r}^{n_{r}}}{n_{1}^{k_{1}}\dotsm
n_{r}^{k_{r}}}\quad\text{and}\quad\operatorname{Li}^{\star}_{{\boldsymbol{\sl{k}}}}({\boldsymbol{\sl{x}}}):=\displaystyle\sum_{n_{1}\geq
n_{2}\geq\cdots\geq n_{r}>0}\displaystyle\frac{x_{1}^{n_{1}}\dotsm
x_{r}^{n_{r}}}{n_{1}^{k_{1}}\dotsm n_{r}^{k_{r}}},$ (1.3)

respectively, which converge if $|x_{1}\cdots x_{j}|<1$ for all
$j=1,\dotsc,r$. For a composition ${\boldsymbol{\sl{k}}}=(k_{1},\ldots,k_{r})$
and positive integer $n$, and $z\in\mathbb{R}$, we define the following
_multiple harmonic-type sums_

$\displaystyle\zeta^{\star}_{n}({\boldsymbol{\sl{k}}};z):=\displaystyle\sum\limits_{n\geq
n_{1}\geq\cdots\geq n_{r}>0}\displaystyle\frac{z^{n_{r}}}{n_{1}^{k_{1}}\cdots
n_{r}^{k_{r}}}.$ (1.4)

The motivation of this paper arises from the results of Mneimneh [8], Campbell
[1], Komatsu-Wang [6] and Gen$\check{\rm c}$ev [4]. In a 2023 _Discrete
Mathematics_ paper, Mneimneh [8] introduced the following remarkable formula
for a binomial sum of harmonic numbers

$\displaystyle\displaystyle\sum_{k=0}^{n}H_{k}\binom{n}{k}p^{k}(1-p)^{n-k}=\displaystyle\sum_{i=1}^{n}\displaystyle\frac{1-(1-p)^{i}}{i}\quad(p\in[0,1]).$
(1.5)

The result in (1.5) is of interest due to how it generalizes the known elegant
formula ([10, 13, 14])

$\displaystyle\displaystyle\sum_{k=0}^{n}H_{k}\binom{n}{k}=2^{n}\left(H_{n}-\displaystyle\sum_{j=1}^{n}\displaystyle\frac{1}{j2^{j}}\right).$

In 2023, Campbell [1] gave two new proofs of (1.5) by using Zeilberger’s
algorithm [11] and beta-type integral formula. Further, in 2024, Komatsu and
Wang [6] extended Mneimneh’s formula to the generalized hyperharmonic numbers.
In particular, they proved that this formula (1.5) can be extended to the
generalized harmonic numbers and established explicit formulas. Gen$\check{\rm
c}$ev [4, Thm. 2.1] established a generalization of Mneimneh summation formula

$\displaystyle\displaystyle\sum_{k=0}^{n}\binom{n}{k}\zeta^{\star}_{k}(r;z)p^{k}(1-p)^{k}=\displaystyle\sum_{n\geq
n_{1}\geq\cdots n_{r}\geq
1}\displaystyle\frac{(1-p)^{n_{1}}\left(\left(1+\displaystyle\frac{zp}{1-p}\right)^{n_{r}}-1\right)}{n_{1}\cdots
n_{r}},$ (1.6)

where $r\in\mathbb{N},p,z\in\mathbb{R}$ and $p\neq 1$. Further, Gen$\check{\rm
c}$ev [4] also mentioned that another type of generalization of the Mneimneh
binomial sums by replacing the generalized harmonic numbers with the multiple
harmonic sums could be considered. Very recently, the current authors [9] used
the method of integrals of natural logarithms to establish a general Mneimneh-
type binomial sums involving Bell numbers and some Mneimneh-type binomial sums
involving (alternating) harmonic numbers.

In this paper, we will use the methods of our previous paper [9] to establish
some new formulas of Mneimneh-type binomial sums involving multiple harmonic-
type sums. The main results are the following four theorems.

######  Theorem 1.1.

For any reals $x,y$ and $z\in(-\infty,1]$ with $x/(x+y)\geq 0$ and
$n,p\in\mathbb{N}$, we have

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p};z)$
$\displaystyle=(-1)^{p-1}(x+y)^{n}\displaystyle\sum_{j=1}^{n}\left(\Big{(}\displaystyle\frac{xz+y}{x+y}\Big{)}^{j}-\Big{(}\displaystyle\frac{y}{x+y}\Big{)}^{j}\right)\left\\{\displaystyle\sum_{i=0}^{p-1}(-1)^{i}\displaystyle\frac{Y_{i}(n)}{i!}\displaystyle\frac{s(j,p-i)}{j!}\right\\}.$
(1.7)

######  Theorem 1.2.

For any reals $x,y$ and $z\in(-\infty,1]$ with $x/(x+y)\geq 0$ and
$n,p,m\in\mathbb{N}$, we have

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p-1},m+1;z)$
$\displaystyle=(-1)^{m+p}(x+y)^{n}\displaystyle\sum_{n\geq j\geq l\geq
1}\left(\displaystyle\frac{y}{x+y}\right)^{j}\left(\left(1+\displaystyle\frac{xz}{y}\right)^{l}-1\right)$
$\displaystyle\quad\times\left\\{\displaystyle\sum_{i=0}^{p-1}(-1)^{i}\displaystyle\frac{Y_{i}(n)}{i!}\displaystyle\frac{s(j,p-i)}{j!}\right\\}\left\\{\displaystyle\sum_{h=0}^{m-1}(-1)^{h}\displaystyle\frac{Y_{h}(j)}{h!}\displaystyle\frac{s(l,m-h)}{l!}\right\\}.$
(1.8)

######  Theorem 1.3.

For any reals $x,y$ with $x/(x+y)\geq 0$ and $n,p,r\in\mathbb{N},\
m\in\mathbb{N}_{0}$, we have

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p-1},m+2,\\{1\\}_{r-1})$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{n\geq n_{1}\geq\cdots\geq
n_{p+m+r}\geq
1}\displaystyle\frac{\left(\displaystyle\frac{y}{x+y}\right)^{n_{p}}\left(\displaystyle\frac{x+y}{y}\right)^{n_{p+m+1}}\left(1-\left(\displaystyle\frac{y}{x+y}\right)^{n_{p+m+r}}\right)}{n_{1}\cdots
n_{p+m+r}}.$ (1.9)

Here $s(n,k)$ and $Y_{k}(n)$ stand for the (unsigned) Stirling numbers of the
first kind and Bell numbers (see Section 2), respectively.

######  Theorem 1.4.

For integer $m\in\mathbb{N}_{0}$ and real $y\in(0,1)$, if
$r\in\mathbb{N}\setminus\\{1\\}$ then

$\displaystyle\zeta^{\star}(m+2,\\{1\\}_{r-1})=\operatorname{Li}^{\star}_{\\{1\\}_{m+r+1}}\big{(}y,\\{1\\}_{m},y^{-1},\\{1\\}_{r-1}\big{)}-\operatorname{Li}^{\star}_{\\{1\\}_{m+r+1}}\big{(}y,\\{1\\}_{m},y^{-1},\\{1\\}_{r-2},y\big{)},$
(1.10)

if $r=1$ then

$\displaystyle\zeta^{\star}(m+2)=\operatorname{Li}^{\star}_{\\{1\\}_{m+2}}\big{(}y,\\{1\\}_{m},y^{-1}\big{)}-\operatorname{Li}^{\star}_{\\{1\\}_{m+2}}\big{(}y,\\{1\\}_{m+1}\big{)}.$
(1.11)

The remainder of this paper is organized as follows. In Section 2, we briefly
introduce the (unsigned) Stirling number of the first kind and Bell
polynomials, and give two relations of Stirling numbers, Bell numbers and
multiple harmonic (star) sums. In Section 3, we first give three lemmas and
prove two propositions. Then we apply these lemmas and propositions obtained
to prove Theorems 1.1-1.3. Further, we present some interesting corollaries
and illustrative cases. Finally, we prove the Theorem 1.4 by applying the
Toeplitz principle. In Section 4, we end this paper with a conjecture of
Mneimneh-type binomial sums involving multiple harmonic star sums.

##  2 Preliminaries

###  2.1 (unsigned) Stirling number of the first kind

Let $s(n,k)$ denote the (unsigned) Stirling number of the first kind, which is
defined by [3, 2]

$\displaystyle
n!x\left({1+x}\right)\left({1+\displaystyle\frac{x}{2}}\right)\cdots\left({1+\displaystyle\frac{x}{n}}\right)=\displaystyle\sum\limits_{k=0}^{n}s(n+1,k+1)x^{k+1},$
(2.12)

with $s(n,k):=0$ if $n<k$, and $s(n,0)=s(0,k):=0,\ s(0,0):=1$, or
equivalently, by the generating function:

$\displaystyle{\log^{k}}({1-x})={\left({-1}\right)^{k}}k!\displaystyle\sum\limits_{n=1}^{\infty}{s(n,k)\displaystyle\frac{{{x^{n}}}}{{n!}}}\quad(x\in[-1,1)).$

The Stirling numbers ${s\left({n,k}\right)}$ of the first kind satisfy a
recurrence relation in the form

$s({n,k})=s({n-1,k-1})+\left({n-1}\right)s({n-1,k})\quad(n,k\in\mathbb{N}).$

Obviously, ${s\left({n,k}\right)}$ can be expressed in terms of a linear
combinations of products of harmonic numbers (see (1.2)). In particular,

$\displaystyle s({n,1})=\left({n-1}\right)!,$ $\displaystyle
s\left({n,2}\right)=\left({n-1}\right)!{H_{n-1}},$ $\displaystyle
s({n,3})=\displaystyle\frac{{\left({n-1}\right)!}}{2}\left[{H_{n-1}^{2}-{H^{(2)}_{n-1}}}\right],$
$\displaystyle
s({n,4})=\displaystyle\frac{{\left({n-1}\right)!}}{6}\left[{H_{n-1}^{3}-3{H_{n-1}}{H^{(2)}_{n-1}}+2{H^{(3)}_{n-1}}}\right],$
$\displaystyle
s({n,5})=\displaystyle\frac{{\left({n-1}\right)!}}{{24}}\left[{H_{n-1}^{4}-6{H^{(4)}_{n-1}}-6H_{n-1}^{2}{H^{(2)}_{n-1}}+3(H^{(2)}_{n-1})^{2}+8H_{n-1}{H^{(3)}_{n-1}}}\right].$

In particular, we proved the following relation ([15, Thm 2.5])

$\displaystyle
s\left({n,k}\right)=\left({n-1}\right)!{\zeta_{n-1}}({{{\left\\{1\right\\}}_{k-1}}})\quad(k,n\in\mathbb{N}).$
(2.13)

###  2.2 Bell polynomials

Define the _exponential partial Bell polynomials_ $B_{n,k}$ by

$\displaystyle\frac{1}{k!}\left(\displaystyle\sum_{n=1}^{\infty}x_{n}\displaystyle\frac{t^{n}}{n!}\right)^{k}=\displaystyle\sum_{n=k}^{\infty}B_{n,k}(x_{1},x_{2},\ldots,x_{n})\displaystyle\frac{t^{n}}{n!}\,,\quad
k=0,1,2,\ldots\,,$

and the _exponential complete Bell polynomials_ $Y_{n}$ by

$Y_{n}(x_{1},x_{2},\ldots,x_{n}):=\displaystyle\sum_{k=0}^{n}B_{n,k}(x_{1},x_{2},\ldots,x_{n})$

(see [2, Section 3.3]). According to [12, Eq. (2.44)], the complete Bell
polynomials $Y_{n}$ satisfy the recurrence

$Y_{0}=1\,,\quad
Y_{n}(x_{1},x_{2},\ldots,x_{n})=\displaystyle\sum_{j=0}^{n-1}\binom{n-1}{j}x_{n-j}Y_{j}(x_{1},x_{2},\ldots,x_{j})\,,\quad
n\geq 1\,,$

from which, the first few polynomials can be obtained immediately:

$\displaystyle Y_{0}=1\,,\quad Y_{1}(x_{1})=x_{1}\,,\quad
Y_{2}(x_{1},x_{2})=x^{2}_{1}+x_{2}\,,\quad
Y_{3}(x_{1},x_{2},x_{3})=x^{3}_{1}+3x_{1}x_{2}+x_{3}\,,$ $\displaystyle
Y_{4}(x_{1},x_{2},x_{3},x_{4})=x^{4}_{1}+6x^{2}_{1}x_{2}+4x_{1}x_{3}+3x^{2}_{2}+x_{4}\,.$

Define the _Bell number_ ${Y_{k}}(n)$ by

$\displaystyle{Y_{k}}(n):={Y_{k}}({{H_{n}},1!{H_{n}^{(2)}},2!{H_{n}^{(3)}},\cdots,({k-1})!{H_{n}^{(k)}}}).$
(2.14)

Clearly, the Bell number ${Y_{k}}(n)$ is a rational linear combination of
products of harmonic numbers. We have

$\displaystyle{Y_{1}}(n)={H_{n}},$
$\displaystyle{Y_{2}}(n)=H_{n}^{2}+{H^{(2)}_{n}},$
$\displaystyle{Y_{3}}(n)=H_{n}^{3}+3{H_{n}}{H^{(2)}_{n}}+2{H^{(3)}_{n}},$
$\displaystyle{Y_{4}}(n)=H_{n}^{4}+8{H_{n}}{H^{(3)}_{n}}+6H_{n}^{2}{H^{(2)}_{n}}+3(H^{(2)}_{n})^{2}+6{H^{(4)}_{n}},$
$\displaystyle{Y_{5}}(n)=H_{n}^{5}+10H_{n}^{3}{H^{(2)}_{n}}+20H_{n}^{2}{H^{(3)}_{n}}+15{H_{n}}({H^{(2)}_{n}})^{2}+30{H_{n}}{H^{(4)}_{n}}+20{H^{(2)}_{n}}{H^{(3)}_{n}}+24{H^{(5)}_{n}}.$

In [15, Eq. (2.9)], we showed

$\displaystyle\zeta_{n}^{\star}\left({{{\\{1\\}}_{m}}}\right)=\displaystyle\frac{1}{{m!}}{Y_{m}}\left(n\right)\quad(n,m\in\mathbb{N}_{0}).$
(2.15)

##  3 Proofs of Main Results

###  3.1 Lemmas and Propositions

Next, we present some lemmas and propositions, which are useful in the
development of our main theorems.

######  Lemma 3.1.

([16, Thm. 2.9] ) For $n\in\mathbb{N}$ and $p\in\mathbb{N}_{0}$, we have

$\displaystyle\displaystyle\int\limits_{0}^{1}{{t^{n-1}}{{\log}^{p}}(1-t)}dt={({-1})^{p}}p!\displaystyle\frac{\zeta_{n}^{\star}(\\{1\\}_{p})}{n}.$
(3.16)

######  Lemma 3.2.

([15, Thm. 2.2]) For $n,p\in\mathbb{N}$ and $x\in(-\infty,1)$, we have

$\displaystyle\displaystyle\int\limits_{0}^{x}{{t^{n-1}}{{\log}^{p}}\left({1-t}\right)}dt=p!\displaystyle\frac{{{{\left({-1}\right)}^{p}}}}{n}\zeta_{n}^{\star}({{{\\{1\\}}_{p}};x})$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad+\displaystyle\frac{1}{n}\displaystyle\sum\limits_{j=0}^{p-1}{{{\left({-1}\right)}^{j}}j!\binom{p}{j}{{\log}^{p-j}}\left({1-x}\right)}\left({\zeta_{n}^{\star}({{{\\{1\\}}_{j}};x})-\zeta_{n}^{\star}({{{\\{1\\}}_{j}}})}\right).$
(3.17)

(Note that in [15, Thm. 2.2], the range of values for $x$ is $x\in[-1,1)$, but
in fact, the above equation also holds for $x\in(-\infty,-1]$.)

######  Lemma 3.3.

([17, Thm. 4.3]) For
${{\boldsymbol{\sl{k}}}}=(k_{1},\ldots,k_{r})\in\mathbb{N}^{r}$ and
$l\in\mathbb{N}_{0},n\in\mathbb{N}$, we have

$\displaystyle\displaystyle\sum_{n\geq n_{1}\geq n_{2}\geq\cdots\geq
n_{r}>0}\prod_{j=1}^{r}\displaystyle\frac{1}{(n_{j}+l)^{k_{j}}}=(-1)^{r}\displaystyle\sum_{j=0}^{r}(-1)^{j}\zeta^{\star}_{n+l}(\overrightarrow{{\boldsymbol{\sl{k}}}}_{\hskip-2.0ptj})\zeta_{l}(\overleftarrow{{\boldsymbol{\sl{k}}}}_{\hskip-2.0ptj+1}),$
(3.18)

where
$\overrightarrow{{\boldsymbol{\sl{k}}}}_{\hskip-2.0ptj}:=(k_{1},\ldots,k_{j})$
and
$\overleftarrow{{\boldsymbol{\sl{k}}}}_{\hskip-2.0ptj}:=(k_{r},\ldots,k_{j})$
for all $1\leq j\leq r$.

######  Proposition 3.4.

For reals $a,z$ with $a\neq 0$ and positive integers $m,j$, we have

$\displaystyle\displaystyle\int_{z>t_{1}>\cdots>t_{m}>0}\displaystyle\frac{(at_{m}+1)^{j}-1}{t_{1}\cdots
t_{m}}dt_{m}\cdots dt_{1}$
$\displaystyle=(-1)^{m-1}\displaystyle\sum_{l=1}^{j}((az+1)^{l}-1)\left\\{\displaystyle\sum_{h=0}^{m-1}(-1)^{h}\displaystyle\frac{Y_{h}(j)}{h!}\displaystyle\frac{s(l,m-h)}{l!}\right\\}.$
(3.19)

###### Proof.

By a direct calculation, one obtains

$\displaystyle\displaystyle\int_{z>t_{1}>\cdots>t_{m}>0}\displaystyle\frac{(at_{m}+1)^{j}-1}{t_{1}\cdots
t_{m}}dt_{m}\cdots dt_{1}$
$\displaystyle=\displaystyle\int_{z>t_{1}>\cdots>t_{m-1}>0}\displaystyle\frac{1}{t_{1}\cdots
t_{m-1}}\left\\{\displaystyle\int_{0}^{t_{m-1}}\displaystyle\frac{(at_{m}+1)^{j}-1}{t_{m}}dt_{m}\right\\}dt_{m-1}\cdots
dt_{1}$
$\displaystyle=\displaystyle\sum_{j_{1}=1}^{j}\displaystyle\frac{1}{j_{1}}\displaystyle\int_{z>t_{1}>\cdots>t_{m-1}>0}\displaystyle\frac{(at_{m-1}+1)^{j_{1}}-1}{t_{1}\cdots
t_{m-1}}dt_{m-1}\cdots dt_{1}=\cdots$
$\displaystyle=\displaystyle\sum_{j_{1}=1}^{j}\displaystyle\frac{1}{j_{1}}\displaystyle\sum_{j_{2}=1}^{j_{1}}\displaystyle\frac{1}{j_{2}}\cdots\displaystyle\sum_{j_{m-1}=1}^{j_{m-2}}\displaystyle\frac{1}{j_{m-1}}\displaystyle\sum_{j_{m}=1}^{j_{m-1}}\displaystyle\frac{(az+1)^{j_{m}}-1}{j_{m}}$
$\displaystyle=\displaystyle\sum_{j\geq j_{1}\geq\cdots\geq j_{m}\geq
1}\displaystyle\frac{(az+1)^{j_{m}}-1}{j_{1}j_{2}\cdots j_{m}}$ (3.20)
$\displaystyle=\displaystyle\sum_{l=1}^{j}\displaystyle\frac{(az+1)^{l}-1}{l}\displaystyle\sum_{j\geq
j_{1}\geq\cdots\geq j_{m-1}\geq l}\displaystyle\frac{1}{j_{1}\cdots j_{m-1}}$
$\displaystyle=\displaystyle\sum_{l=1}^{j}\displaystyle\frac{(az+1)^{l}-1}{l}\displaystyle\sum_{j-l+1\geq
i_{1}\geq\cdots\geq i_{m-1}\geq
1}\displaystyle\frac{1}{(i_{1}+l-1)\cdots(i_{m-1}+l-1)}.$

In Lemma 3.3, replacing $n$ by $j-l+1$ and $l$ by $l-1$, and letting $r=m-1$,
$(k_{1},\ldots,k_{r})=(\\{1\\}_{m-1})$, we easily obtain

$\displaystyle\displaystyle\sum_{j-l+1\geq i_{1}\geq\cdots\geq i_{m-1}\geq
1}\displaystyle\frac{1}{(i_{1}+l-1)\cdots(i_{m-1}+l-1)}$
$\displaystyle\quad\quad=(-1)^{m-1}\displaystyle\sum_{h=0}^{m-1}(-1)^{h}\zeta^{\star}_{j}(\\{1\\}_{h})\zeta_{l-1}(\\{1\\}_{m-1-h}).$
(3.21)

Applying (2.13), (2.15) and (3.1) yields the desired evaluation. ∎

######  Proposition 3.5.

For positive integer $j$ and real $\alpha\neq 0$, we have

$\displaystyle\displaystyle\int_{0}^{1}\log^{m}(1-t)(1+\alpha
t)^{j-1}dt=(-1)^{m+1}m!\displaystyle\frac{(\alpha+1)^{j}}{\alpha
j}\displaystyle\sum_{j\geq j_{1}\geq\cdots\geq j_{m}\geq
1}\displaystyle\frac{(\alpha+1)^{-j_{m}}-1}{j_{1}\cdots j_{m}}.$ (3.22)

###### Proof.

In (3.4), letting $z=1$ and noting the fact that

$\displaystyle\displaystyle\int_{1>t_{1}>\cdots>t_{m-1}>t}\displaystyle\frac{1}{t_{1}\cdots
t_{m-1}}dt_{m-1}\cdots
dt_{1}=\displaystyle\frac{(-1)^{m-1}}{(m-1)!}\log^{m-1}(t),$

the left hand side of (3.4) can be rewritten as

$\displaystyle\displaystyle\int_{1>t_{1}>\cdots>t_{m}>0}\displaystyle\frac{(at_{m}+1)^{j}-1}{t_{1}\cdots
t_{m}}dt_{m}\cdots dt_{1}$
$\displaystyle=\displaystyle\int_{0}^{1}\displaystyle\frac{(at+1)^{j}-1}{t}\left\\{\displaystyle\int_{1>t_{1}>\cdots>t_{m-1}>t}\displaystyle\frac{1}{t_{1}\cdots
t_{m-1}}dt_{m-1}\cdots dt_{1}\right\\}dt$
$\displaystyle=\displaystyle\frac{(-1)^{m-1}}{(m-1)!}\displaystyle\int_{0}^{1}\displaystyle\frac{\log^{m-1}(t)\Big{(}(at+1)^{j}-1\Big{)}}{t}dt$
$\displaystyle=\displaystyle\frac{(-1)^{m}}{m!}aj\displaystyle\int_{0}^{1}\log^{m}(t)(at+1)^{j-1}dt\quad\quad(t\rightarrow
1-t)$
$\displaystyle=\displaystyle\frac{(-1)^{m}}{m!}aj\displaystyle\int_{0}^{1}\log^{m}(1-t)(a(1-t)+1)^{j-1}dt$
$\displaystyle=\displaystyle\frac{(-1)^{m}}{m!}a(a+1)^{j-1}j\displaystyle\int_{0}^{1}\log^{m}(1-t)\left(1-\displaystyle\frac{a}{a+1}\right)^{j-1}dt.$

Using (3.20) with $z=1$ gives

$\displaystyle\displaystyle\int_{0}^{1}\log^{m}(1-t)\left(1-\displaystyle\frac{a}{a+1}t\right)^{j-1}dt$
$\displaystyle=(-1)^{m}m!\displaystyle\frac{1}{a(a+1)^{j-1}j}\displaystyle\sum_{j\geq
j_{1}\geq\cdots\geq j_{m}\geq
1}\displaystyle\frac{(a+1)^{j_{m}}-1}{j_{1}j_{2}\cdots j_{m}}.$

Finally, setting $\alpha=-a/(a+1)\ (a\neq 0,-1)$, by an elementary
calculation, we obtain the desired evaluation. ∎

###  3.2 Proof of Theorems 1.1, 1.2 and 1.3

From Lemmas 3.1 and 3.2, by a simple calculation, we arrive at the result

$\displaystyle\zeta^{\star}_{n}(\\{1\\}_{p};z)-\zeta^{\star}_{n}(\\{1\\}_{p})=\displaystyle\frac{n}{p!}\displaystyle\sum_{j=0}^{p}(-1)^{j}\binom{p}{j}\log^{p-j}(1-z)\displaystyle\int_{1}^{z}t^{n-1}\log^{j}(1-t)dt.$
(3.23)

Hence,

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\left(\zeta_{k}^{\star}(\\{1\\}_{p};z)-\zeta_{k}^{\star}(\\{1\\}_{p}\right)$
$\displaystyle=\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}k\displaystyle\sum_{j=0}^{p}\displaystyle\frac{(-1)^{j}}{j!}\displaystyle\frac{\log^{p-j}(1-z)}{(p-j)!}\displaystyle\int_{1}^{z}t^{k-1}\log^{j}(1-t)dt$
$\displaystyle=nx\displaystyle\sum_{j=0}^{p}\displaystyle\frac{(-1)^{j}}{j!}\displaystyle\frac{\log^{p-j}(1-z)}{(p-j)!}\displaystyle\int_{1}^{z}\log^{j}(1-t)(xt+y)^{n-1}dt\quad\quad\quad(\text{letting}\
u=tx+y)$
$\displaystyle=n\displaystyle\sum_{j=0}^{p}\displaystyle\frac{(-1)^{j}}{j!}\displaystyle\frac{\log^{p-j}(1-z)}{(p-j)!}\displaystyle\int_{x+y}^{xz+y}u^{n-1}\log^{j}\Big{(}\displaystyle\frac{x+y-u}{x}\Big{)}du$
$\displaystyle=n\displaystyle\sum_{j=0}^{p}\displaystyle\frac{(-1)^{j}}{j!}\displaystyle\frac{\log^{p-j}(1-z)}{(p-j)!}\displaystyle\int_{x+y}^{xz+y}u^{n-1}\left\\{\log\Big{(}\displaystyle\frac{x+y}{x}\Big{)}+\log\Big{(}1-\displaystyle\frac{u}{x+y}\Big{)}\right\\}^{j}du$
$\displaystyle=n\displaystyle\sum_{j=0}^{p}\displaystyle\frac{(-1)^{j}}{j!}\displaystyle\frac{\log^{p-j}(1-z)}{(p-j)!}\displaystyle\sum_{i=0}^{j}\binom{j}{i}\log^{j-i}\Big{(}\displaystyle\frac{x+y}{x}\Big{)}\displaystyle\int_{x+y}^{xz+y}u^{n-1}\log^{i}\Big{(}1-\displaystyle\frac{u}{x+y}\Big{)}du$
$\displaystyle=n(x+y)^{n}\displaystyle\sum_{j=0}^{p}\displaystyle\frac{(-1)^{j}}{j!}\displaystyle\frac{\log^{p-j}(1-z)}{(p-j)!}\displaystyle\sum_{i=0}^{j}\binom{j}{i}\log^{j-i}\Big{(}\displaystyle\frac{x+y}{x}\Big{)}\displaystyle\int_{1}^{(xz+y)(x+y)^{-1}}v^{n-1}\log^{i}(1-v)dv.$

Applying (3.16) and (3.2) gives

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\left(\zeta_{k}^{\star}(\\{1\\}_{p};z)-\zeta_{k}^{\star}(\\{1\\}_{p}\right)$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{j=0}^{p}\displaystyle\frac{(-1)^{j}}{j!}\displaystyle\frac{\log^{p-j}(1-z)}{(p-j)!}\displaystyle\sum_{i=0}^{j}\binom{j}{i}\log^{j-i}\Big{(}\displaystyle\frac{x+y}{x}\Big{)}$
$\displaystyle\quad\quad\quad\quad\quad\times\displaystyle\sum_{l=0}^{i}(-1)^{l}l!\binom{i}{l}\log^{i-l}\Big{(}\displaystyle\frac{(1-z)x}{x+y}\Big{)}\left(\zeta^{\star}_{n}\Big{(}\\{1\\}_{l};\displaystyle\frac{xz+y}{x+y}\Big{)}-\zeta^{\star}_{n}(\\{1\\}_{l})\right)$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{p\geq j\geq i\geq l\geq
0}(-1)^{j+l}\displaystyle\frac{\log^{p-j}(1-z)\log^{j-i}\Big{(}\displaystyle\frac{x+y}{x}\Big{)}\log^{i-l}\Big{(}\displaystyle\frac{(1-z)x}{x+y}\Big{)}}{(p-j)!(j-i)!(i-l)!}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\left(\zeta^{\star}_{n}\Big{(}\\{1\\}_{l};\displaystyle\frac{xz+y}{x+y}\Big{)}-\zeta^{\star}_{n}(\\{1\\}_{l})\right)$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{l+h+g+q=p,\atop l,h,g,q\geq
0}(-1)^{h+g}\displaystyle\frac{\log^{q}(1-z)\log^{g}\Big{(}\displaystyle\frac{x+y}{x}\Big{)}\log^{h}\Big{(}\displaystyle\frac{(1-z)x}{x+y}\Big{)}}{q!g!h!}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\left(\zeta^{\star}_{n}\Big{(}\\{1\\}_{l};\displaystyle\frac{xz+y}{x+y}\Big{)}-\zeta^{\star}_{n}(\\{1\\}_{l})\right)$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{l=0}^{p}\left(\zeta^{\star}_{n}\Big{(}\\{1\\}_{l};\displaystyle\frac{xz+y}{x+y}\Big{)}-\zeta^{\star}_{n}(\\{1\\}_{l})\right)$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\times\displaystyle\sum_{h+g+q=p-l,\atop
h,g,q\geq
0}(-1)^{h+g}\displaystyle\frac{\log^{q}(1-z)\log^{g}\Big{(}\displaystyle\frac{x+y}{x}\Big{)}\log^{h}\Big{(}\displaystyle\frac{(1-z)x}{x+y}\Big{)}}{q!g!h!}$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{l=0}^{p}\left(\zeta^{\star}_{n}\Big{(}\\{1\\}_{l};\displaystyle\frac{xz+y}{x+y}\Big{)}-\zeta^{\star}_{n}(\\{1\\}_{l})\right)$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\times\displaystyle\frac{1}{(p-l)!}\left\\{\log(1-z)+\log\Big{(}\displaystyle\frac{x}{x+y}\Big{)}+\log\Big{(}\displaystyle\frac{x+y}{(1-z)x}\Big{)}\right\\}^{p-l}$
$\displaystyle=(x+y)^{n}\left(\zeta^{\star}_{n}\Big{(}\\{1\\}_{p};\displaystyle\frac{xz+y}{x+y}\Big{)}-\zeta^{\star}_{n}(\\{1\\}_{p})\right).$

Clearly, if setting $z=0$ then

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p})=(x+y)^{n}\left(\zeta^{\star}_{n}(\\{1\\}_{p})-\zeta^{\star}_{n}\Big{(}\\{1\\}_{p};\displaystyle\frac{y}{x+y}\Big{)}\right).$

Therefore, we have

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p};z)$
$\displaystyle=(x+y)^{n}\left(\zeta^{\star}_{n}\Big{(}\\{1\\}_{p};\displaystyle\frac{xz+y}{x+y}\Big{)}-\zeta^{\star}_{n}\Big{(}\\{1\\}_{p};\displaystyle\frac{y}{x+y}\Big{)}\right)$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{n\geq n_{1}\geq\cdots\geq n_{p}\geq
1}\displaystyle\frac{\Big{(}\displaystyle\frac{xz+y}{x+y}\Big{)}^{n_{p}}-\Big{(}\displaystyle\frac{y}{x+y}\Big{)}^{n_{p}}}{n_{1}\cdots
n_{p}}$ (3.24)
$\displaystyle=(x+y)^{n}\displaystyle\sum_{j=1}^{n}\left(\Big{(}\displaystyle\frac{xz+y}{x+y}\Big{)}^{j}-\Big{(}\displaystyle\frac{y}{x+y}\Big{)}^{j}\right)\displaystyle\sum_{n\geq
n_{1}\geq\cdots\geq n_{p-1}\geq j}\displaystyle\frac{1}{n_{1}\cdots n_{p-1}}.$

Finally, applying (2.13), (2.15) and (3.1) yields the desired evaluation.
Thus, this concludes the proof of Theorem 1.1.

Replacing $z$ by $t_{m}$ in (1.1), then applying
$\displaystyle\int_{z>t_{1}>\cdots>t_{m}>0}\displaystyle\frac{(\cdot)}{t_{1}\cdots
t_{m}}dt_{m}\cdots dt_{1}$ to it and using Proposition 3.4, we deduce

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p-1},m+1;z)$
$\displaystyle=(-1)^{p-1}(x+y)^{n}\displaystyle\sum_{j=1}^{n}\left\\{\displaystyle\sum_{i=0}^{p-1}(-1)^{i}\displaystyle\frac{Y_{i}(n)}{i!}\displaystyle\frac{s(j,p-i)}{j!}\right\\}(x+y)^{-j}y^{j}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\times\displaystyle\int_{z>t_{1}>\cdots>t_{m}>0}\displaystyle\frac{\left(\displaystyle\frac{x}{y}t_{m}+1\right)^{j}-1}{t_{1}\cdots
t_{m}}dt_{m}\cdots dt_{1}$
$\displaystyle=(-1)^{p+m}(x+y)^{n}\displaystyle\sum_{j=1}^{n}\left\\{\displaystyle\sum_{i=0}^{p-1}(-1)^{i}\displaystyle\frac{Y_{i}(n)}{i!}\displaystyle\frac{s(j,p-i)}{j!}\right\\}(x+y)^{-j}y^{j}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\times\displaystyle\sum_{l=1}^{j}\left(\Big{(}\displaystyle\frac{xz}{y}+1\Big{)}^{l}-1\right)\left\\{\displaystyle\sum_{h=0}^{m-1}(-1)^{h}\displaystyle\frac{Y_{h}(j)}{h!}\displaystyle\frac{s(l,m-h)}{l!}\right\\}.$

Thus, this completes the proof of Theorem 1.2.

On the other hand, applying
$\displaystyle\int_{z>t_{1}>\cdots>t_{m}>0}\displaystyle\frac{(\cdot)}{t_{1}\cdots
t_{m}}dt_{m}\cdots dt_{1}$ to (3.2) and using (3.20), we obtain another form
of (1.2) ($p\in\mathbb{N},\ m\in\mathbb{N}_{0}$)

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p-1},m+1;z)$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{n\geq n_{1}\geq\cdots\geq
n_{p+m}\geq
1}\displaystyle\frac{(x+y)^{-n_{p}}y^{n_{p}-n_{p+m}}\left((xz+y)^{n_{p+m}}-y^{n_{p+m}}\right)}{n_{1}\cdots
n_{p+m}}.$ (3.25)

Multiplying (3.2) by $\log^{r-1}(1-z)/z$ and integrating over the interval
(0,1), and applying (3.16), we have

$\displaystyle(-1)^{r-1}(r-1)!\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p-1},m+1;z)$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{n\geq n_{1}\geq\cdots\geq
n_{p+m}\geq 1}\displaystyle\frac{(x+y)^{-n_{p}}y^{n_{p}}}{n_{1}\cdots
n_{p+m}}\displaystyle\int_{0}^{1}\displaystyle\frac{\log^{r-1}(1-z)\left(\Big{(}\displaystyle\frac{xz}{y}+1\Big{)}^{n_{p+m}}-1\right)}{z}dz$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{n\geq n_{1}\geq\cdots\geq
n_{p+m}\geq 1}\displaystyle\frac{(x+y)^{-n_{p}}y^{n_{p}}}{n_{1}\cdots
n_{p+m}}\displaystyle\sum_{n_{p+m+1}=1}^{n_{p+m}}\displaystyle\frac{x}{y}\displaystyle\int_{0}^{1}\log^{r-1}(1-z)\Big{(}\displaystyle\frac{xz}{y}+1\Big{)}^{n_{p+m+1}-1}dz.$
(3.26)

Finally, using Proposition 3.5, by a straight-forward calculation, we obtain
(1.3). Thus, this concludes the proof of the Theorem 1.3.

###  3.3 Corollaries and Examples

From Theorem 1.2, letting $m=1$ or $p=1$ give the following corollaries.

######  Corollary 3.6.

For any reals $x,y$ and $z\in(-\infty,1]$ with $x/(x+y)\geq 0$ and
$n,p\in\mathbb{N}$, we have

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(\\{1\\}_{p-1},2;z)$
$\displaystyle=(-1)^{p-1}(x+y)^{n}\displaystyle\sum_{n\geq j\geq l\geq
1}\displaystyle\frac{1}{l}\left(\displaystyle\frac{y}{x+y}\right)^{j}\left(\left(1+\displaystyle\frac{xz}{y}\right)^{l}-1\right)\left\\{\displaystyle\sum_{i=0}^{p-1}(-1)^{i}\displaystyle\frac{Y_{i}(n)}{i!}\displaystyle\frac{s(j,p-i)}{j!}\right\\}.$
(3.27)

######  Corollary 3.7.

For any reals $x,y$ and $z\in(-\infty,1]$ with $x/(x+y)\geq 0$ and
$n,m\in\mathbb{N}$, we have

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(m+1;z)$
$\displaystyle=(-1)^{m-1}(x+y)^{n}\displaystyle\sum_{n\geq j\geq l\geq
1}\displaystyle\frac{1}{j}\left(\displaystyle\frac{y}{x+y}\right)^{j}\left(\left(1+\displaystyle\frac{xz}{y}\right)^{l}-1\right)\left\\{\displaystyle\sum_{h=0}^{m-1}(-1)^{h}\displaystyle\frac{Y_{h}(j)}{h!}\displaystyle\frac{s(l,m-h)}{l!}\right\\}.$
(3.28)

Further, from Theorems 1.1-1.3, and Corollaries 3.6-3.7, we obtain the
following specific cases.

######  Example 3.8.

For any reals $x,y$ with $x/(x+y)\geq 0$,

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}{\bar{H}}^{(2)}_{k}=\displaystyle\sum_{n\geq
j\geq l\geq
1}\displaystyle\frac{1}{jl}(x+y)^{n-j}y^{j-l}\Big{(}y^{l}-(y-x)^{l}\Big{)},$
$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}{H}^{(3)}_{k}=\displaystyle\sum_{n\geq
j\geq l\geq
1}\displaystyle\frac{1}{jl}(x+y)^{n-j}y^{j-l}\Big{(}(x+y)^{l}-y^{l}\Big{)}(H_{j}-H_{l-1}),$
$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta^{\star}_{k}(1,2)=\displaystyle\sum_{n\geq
j\geq l\geq
1}\displaystyle\frac{1}{jl}(x+y)^{n-j}y^{j-l}\Big{(}(x+y)^{l}-y^{l}\Big{)}(H_{n}-H_{j-1}),$
$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta^{\star}_{k}(2,1)=\displaystyle\sum_{n\geq
j\geq l\geq h\geq
1}\displaystyle\frac{1}{jlh}(x+y)^{n-j}y^{j-l}(x+y)^{l}\Big{(}1-(x+y)^{-h}y^{h}\Big{)},$

where ${\bar{H}}^{(2)}_{k}$ is the _$k$ -th alternating harmonic number of
order 2_ defined by

$\displaystyle{\bar{H}}^{(2)}_{k}:=\displaystyle\sum_{j=1}^{k}\displaystyle\frac{(-1)^{j-1}}{j^{2}}.$

More generally, the _$k$ -th alternating harmonic number of order p_
${\bar{H}}^{(p)}_{k}\ (p\in\mathbb{N})$ is defined by

$\displaystyle{\bar{H}}^{(p)}_{k}:=\displaystyle\sum_{j=1}^{k}\displaystyle\frac{(-1)^{j-1}}{j^{p}}\quad\text{and}\quad{\bar{H}}^{(p)}_{0}:=0.$

Noting the fact that

$H_{k}H_{k}^{(2)}=\zeta^{\star}_{k}(1,2)+\zeta^{\star}_{k}(2,1)-H_{k}^{(3)},$

we get the Mneimneh-type binomial sums involving two harmonic numbers

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}H_{k}H_{k}^{(2)}$
$\displaystyle=\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta^{\star}_{k}(1,2)+\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta^{\star}_{k}(2,1)-\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}{H}^{(3)}_{k}$
$\displaystyle=\displaystyle\sum_{n\geq j\geq l\geq
1}\displaystyle\frac{1}{jl}(x+y)^{n-j}y^{j-l}\Big{(}(x+y)^{l}-y^{l}\Big{)}\Big{(}H_{n}-2H_{j}+H_{l}+\displaystyle\frac{1}{j}-\displaystyle\frac{1}{l}\Big{)}$
$\displaystyle\quad+\displaystyle\sum_{n\geq j\geq l\geq
1}\displaystyle\frac{1}{jl}(x+y)^{n-j}y^{j-l}(x+y)^{l}\left(H_{l}-\displaystyle\sum_{h=1}^{l}\displaystyle\frac{(x+y)^{-h}y^{h}}{h}\right).$
(3.29)

On the other hand, from [9, Thm. 1.1], setting $p=$3 yields

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\Big{(}H_{k}^{3}+3H_{k}H_{k}^{(2)}+2H_{k}^{(3)}\Big{)}$
$\displaystyle=3(x+y)^{n}\displaystyle\sum_{j=0}^{n}\displaystyle\frac{1}{j}\left(1-\Big{(}\displaystyle\frac{y}{x+y}\Big{)}^{j}\right)\left\\{H_{n}^{2}+H_{n}^{(2)}-2H_{n}H_{j-1}+H_{j-1}^{2}-H_{j-1}^{(2)}\right\\}.$
(3.30)

Hence, by an elementary calculation, we deduce the Mneimneh-type binomial sums
involving cubic harmonic numbers

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}H_{k}^{3}$
$\displaystyle=3(x+y)^{n}\displaystyle\sum_{j=0}^{n}\displaystyle\frac{1}{j}\left(1-\Big{(}\displaystyle\frac{y}{x+y}\Big{)}^{j}\right)\left\\{H_{n}^{2}+H_{n}^{(2)}-2H_{n}H_{j-1}+H_{j-1}^{2}-H_{j-1}^{(2)}\right\\}$
$\displaystyle\quad-\displaystyle\sum_{n\geq j\geq l\geq
1}\displaystyle\frac{1}{jl}(x+y)^{n-j}y^{j-l}\Big{(}(x+y)^{l}-y^{l}\Big{)}\Big{(}3H_{n}-4H_{j}+H_{l}+\displaystyle\frac{3}{j}-\displaystyle\frac{1}{l}\Big{)}$
$\displaystyle\quad-3\displaystyle\sum_{n\geq j\geq l\geq
1}\displaystyle\frac{1}{jl}(x+y)^{n-j}y^{j-l}(x+y)^{l}\left(H_{l}-\displaystyle\sum_{h=1}^{l}\displaystyle\frac{(x+y)^{-h}y^{h}}{h}\right).$
(3.31)

###  3.4 Proof of Theorem 1.4

The proof of Theorem 1.4 is based on the following well-known Toeplitz limit
theorem.

######  Lemma 3.9.

([7, Thm. 4]) Assume that $\displaystyle\lim_{n\rightarrow\infty}x_{n}=0$ and
the terms $a_{nk}$ form a triangular system

$\displaystyle\left({\begin{array}[]{*{20}{c}}{{a_{00}}}&0&0&\cdots&0\\\
{{a_{10}}}&{{a_{11}}}&0&\cdots&0\\\
{{a_{20}}}&{{a_{21}}}&{{a_{22}}}&\cdots&0\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
{{a_{n0}}}&{{a_{n1}}}&{{a_{n2}}}&\cdots&{{a_{nn}}}\end{array}}\right)$

and satisfy the following conditions: (a) every column of the system contains
a null sequence, i.e. for fixed $k\geq 0$,
$\displaystyle\lim_{n\rightarrow\infty}a_{nk}=0$, (b) there exists a constant
$K$ independent of $n$ such that $\displaystyle\sum_{k=0}^{n}|a_{nk}|<K$. Then

$\displaystyle\displaystyle\lim_{n\rightarrow\infty}\displaystyle\sum_{k=0}^{n}a_{nk}x_{k}=0.$
(3.32)

Proof of Theorem 1.4. We only prove that Theorem 1.4 holds for $r\geq 2$. From
Theorem 1.3, setting $p=1,\ x+y=1$ and $x\in[0,1)$ gives

$\displaystyle\displaystyle\sum_{k=0}^{n}(1-y)^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(m+2,\\{1\\}_{r-1})=\displaystyle\sum_{n\geq
n_{1}\geq\cdots\geq n_{m+r+1}\geq
1}\displaystyle\frac{y^{n_{1}-n_{m+2}}(1-y^{n_{m+r+1}})}{n_{1}\cdots
n_{m+r+1}}.$ (3.33)

By the relation in (3.33), letting $n\rightarrow\infty$, we immediately
discover that for $y\neq 0$

$\displaystyle\displaystyle\lim_{n\rightarrow\infty}\displaystyle\sum_{k=0}^{n}(1-y)^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(m+2,\\{1\\}_{r-1})$
$\displaystyle=\operatorname{Li}^{\star}_{\\{1\\}_{m+r+1}}\big{(}y,\\{1\\}_{m},y^{-1},\\{1\\}_{r-1}\big{)}-\operatorname{Li}^{\star}_{\\{1\\}_{m+r+1}}\big{(}y,\\{1\\}_{m},y^{-1},\\{1\\}_{r-2},y\big{)}.$
(3.34)

Therefore, for the proof of (1.10), it suffices to show that

$\displaystyle\displaystyle\lim_{n\rightarrow\infty}\left(\zeta^{\star}(m+2,\\{1\\}_{r-1})-\displaystyle\sum_{k=0}^{n}(1-y)^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(m+2,\\{1\\}_{r-1})\right)=0,$
(3.35)

where $m,r,n$ and $y$ satisfy the assumptions of Theorem 1.4.

Noting the fact that

$\displaystyle\sum_{k=0}^{n}(1-y)^{k}y^{n-k}\binom{n}{k}=1,$

we have

$\displaystyle\zeta^{\star}(m+2,\\{1\\}_{r-1})-\displaystyle\sum_{k=0}^{n}(1-y)^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}(m+2,\\{1\\}_{r-1})$
$\displaystyle=\displaystyle\sum_{k=0}^{n}(1-y)^{k}y^{n-k}\binom{n}{k}\left(\zeta^{\star}(m+2,\\{1\\}_{r-1})-\zeta_{k}^{\star}(m+2,\\{1\\}_{r-1})\right).$
(3.36)

Now, set for every $k,n\in\mathbb{N}$,

$\displaystyle a_{nk}:=\binom{n}{k}(1-y)^{k}y^{n-k}\quad\text{and}\quad
x_{k}:=\zeta^{\star}(m+2,\\{1\\}_{r-1})-\zeta_{k}^{\star}(m+2,\\{1\\}_{r-1})$

and let us verify the assumptions of Lemma 3.9. Clearly, the condition
$\displaystyle\lim_{k\rightarrow\infty}x_{k}=0$ and condition (a)
$\displaystyle\lim_{n\rightarrow\infty}a_{nk}=0$ are satisfied (since
$y\in(0,1)$). To fulfill condition (b), it is necessary that
$\displaystyle\sum_{k=0}^{n}|a_{nk}|$ is bounded. Noticing that

$\displaystyle\displaystyle\sum_{k=0}^{n}|a_{nk}|=\displaystyle\sum_{k=0}^{n}\binom{n}{k}|(1-y)^{k}y^{n-k}|=(1-y+y)^{n}=1,$
(3.37)

and applying Lemma 3.9, we obtain the desired evaluation (1.10). Similarly, we
also prove (1.11). This concludes the proof of Theorem 1.4.

##  4 A Conjecture

For any ${{\boldsymbol{\sl{k}}}}:=(k_{1},k_{2},\ldots,\ k_{r})$, we put

$\displaystyle|{{\boldsymbol{\sl{k}}}}|_{j}:=k_{1}+k_{2}+\cdots+k_{j}\quad\text{and}\quad|{{\boldsymbol{\sl{k}}}}|_{0}:=0.$

Now, we end this paper by the following conjecture.

######  Conjecture 4.1.

Let $r\in\mathbb{N}_{0}$ and $n\in\mathbb{N}$. For any reals $x,y$ and ${\bf
p}_{r+1}:=(p_{1},p_{2},\ldots,\ p_{r+1})\in\mathbb{N}^{r+1},\ {\bf
m}_{r}:=(m_{1},\ldots,m_{r})\in\mathbb{N}_{0}^{r}$, we have

$\displaystyle\displaystyle\sum_{k=0}^{n}x^{k}y^{n-k}\binom{n}{k}\zeta_{k}^{\star}\left(\\{1\\}_{p_{1}-1},m_{1}+2,\ldots,\\{1\\}_{p_{r}-1},m_{r}+2,\\{1\\}_{p_{r+1}-1}\right)$
$\displaystyle=(x+y)^{n}\displaystyle\sum_{n\geq n_{1}\geq\cdots\geq n_{|{\bf
p}|_{r+1}+|{\bf m}|_{r}+r-1}\geq
1}\displaystyle\frac{\left(\displaystyle\frac{y}{x+y}\right)^{\displaystyle\sum_{j=1}^{r}\Big{(}n_{|{\bf
p}|_{j}+|{\bf m}|_{j-1}+j-1}-n_{|{\bf p}|_{j}+|{\bf
m}|_{j}+j}\Big{)}}}{n_{1}\cdots n_{|{\bf p}|_{r+1}+|{\bf m}|_{r}+r-1}}$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\times\left(1-\left(\displaystyle\frac{y}{x+y}\right)^{n_{|{\bf
p}|_{r+1}+|{\bf m}|_{r}+r-1}}\right).$ (4.38)

######  Remark 4.2.

Applying the Toeplitz principle, we also deduce the corresponding
(conjectural) functional multiple polylogarithmic identity similar to Theorem
1.4.

Declaration of competing interest. The authors declares that they has no known
competing financial interests or personal relationships that could have
appeared to influence the work reported in this paper.

Data availability. No data was used for the research described in the article.

Acknowledgments. The authors are grateful to Prof. M. Gen$\check{\rm c}$ev for
his opinion concerning the relation in Theorem 1.4. Ce Xu is supported by the
National Natural Science Foundation of China (Grant No. 12101008), the Natural
Science Foundation of Anhui Province (Grant No. 2108085QA01) and the
University Natural Science Research Project of Anhui Province (Grant No.
KJ2020A0057).

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