<s>[INST] Calculate the limit of the following recurrent series Find the limit of the series : $\ q_{n+1,}=q_n + \frac{2}{q_n}$ with $q_0=1,n\ge 0$.
 [/INST] I will show that
$q_n$ is unbounded
and that
$\frac12+\sqrt{2n-\frac74}
\lt q_n
<3+2\sqrt{ 2 n+\frac14}
$.
Moreover,
if,
as these inequalities make plausible,
$\lim_{n \to \infty} \dfrac{q_n}{\sqrt{n}}
$
exists and this limit is $c$,
then $c = 2$,
so that
$\lim_{n \to \infty} \dfrac{q_n}{\sqrt{n}}
=2
$.
Here we go.
$q_n > 1$ for $n > 1$
and $q_n$ is increasing.
If $q_n$ is bounded,
there is an $L$
such that $q_n \le L$
for all $n$.
Writing
$q_{n+1}=q_n + \frac{2}{q_n}
$
as
$q_{n+1}-q_n 
= \frac{2}{q_n}
$
and summing
$\begin{array}\\
q_{n+1}-q_1
&=\sum_{k=1}^n (q_{k+1}-q_k)\\
&=\sum_{k=1}^n \frac{2}{q_k}\\
&\ge\sum_{k=1}^n \frac{2}{L}\\
&\ge \dfrac{2n}{L}\\
\end{array}
$
which is unbounded,
a contradiction.
Note that this can be used,
as Will Jagy wrote,
to show that
if
$q_{n+1} = q_n+f(q_n)$
where
$f(x) > 0$
and
$f'(x) < 0$
then
$q_n$ is unbounded.-
This can be used
to get explicit bounds.
$\begin{array}\\
q_{n+1}-q_1
&=\sum_{k=1}^n (q_{k+1}-q_k)\\
&=\sum_{k=1}^n \frac{2}{q_k}\\
&\ge\sum_{k=1}^n \frac{2}{q_n}\\
&\ge \dfrac{2n}{q_n}\\
&\gt \dfrac{2n}{q_{n+1}}\\
\end{array}
$
so,
letting
$\frac{q_1}{2} = a$,
$\begin{array}\\
q_n > \dfrac{2n-2}{q_n}+2a\\
\text{or}\\
q_n^2-2aq_n > 2n-2\\
\text{or}\\
q_n^2-2aq_n+a^2 > 2n-2+a^2\\
\text{or}\\
(q_n-a)^2 > 2n-2+a+a^2\\
\text{or}\\
q_n
\gt a+\sqrt{2n-2+a^2}\\
\end{array}
$
Therefore
$q_n
\gt a+\sqrt{2n-2+a^2}\\
= \frac12+\sqrt{2n-\frac74}\\
$.
With this lower bound for $q_n$,
we can now get an upper bound.
Let $2-a^2 = b$.
$\begin{array}\\
q_{n+1}-q_1
&=\sum_{k=1}^n \frac{2}{q_k}\\
&=\frac{2}{q_1}+\sum_{k=2}^n \frac{2}{q_k}\\
&\lt\frac{2}{q_1}+2\sum_{k=2}^n \frac{1}{a+\sqrt{2n-b}}\\
&\lt\frac{2}{q_1}+2\sum_{k=2}^n \int_{k-1}^k\frac{dx}{a+\sqrt{2x-b}}\\
&\lt\frac{2}{q_1}+2\int_{1}^n\frac{dx}{a+\sqrt{2x-b}}\\
&=\frac{2}{q_1}++2(\sqrt{ 2 x-b} - a \log(\sqrt{ 2 x-b} + a))|_{1}^n\\
&=\frac{2}{q_1}+2(\sqrt{2 n-b} - a \log(\sqrt{2 n-b} + a))\\
&\qquad-2(\sqrt{a^2+2}-\log(\sqrt{a^2+2}+a))\\
&=\frac{2}{q_1}+2(\sqrt{a^2 + 2 (n+1)} - a \log(\sqrt{a^2 + 2( n+1)} + a))\\
&\qquad-2(\sqrt{2-b}-\log(\sqrt{2-b}+a))\\
\text{or}\\
q_{n}-q_1
&<\frac{2}{q_1}+2(\sqrt{a^2 + 2n} - a \log(\sqrt{a^2 + 2n} + a))\\
&\qquad-2(\sqrt{a^2+2}-\log(\sqrt{a^2+2}+a))\\
&<\frac{2}{q_1}+2\sqrt{a^2+ 2 n}\\
\end{array}
$
Therefore
$q_n
\lt 2a+\frac{2}{q_1}+2\sqrt{a^2+ 2 n}
= 3+2\sqrt{ 2 n+\frac14}
$.
Assuming a little more,
if
$q_n \approx c\sqrt{n}+O(1)$,
then
$c\sqrt{n+1}-c\sqrt{n}+O(1)
=\dfrac{2}{c\sqrt{n}+O(1)}
=\dfrac{2}{c\sqrt{n}}+O(\dfrac1{n})
$
so, since
$\begin{array}\\
\sqrt{n+1}-\sqrt{n}
&=\dfrac1{2\sqrt{n+d}}
\qquad 0 \le d \le 1, mvt\\
&=\dfrac1{2\sqrt{n}\sqrt{1+d/n}}\\
&=\dfrac1{2\sqrt{n}}(1+O(\dfrac1{n}))\\
\end{array}
$
$\dfrac{c}{2\sqrt{n}}+O(\dfrac1{n^{3/2}})
=\dfrac{2}{c\sqrt{n}}+O(1/n)
$
so
$\dfrac{c}{2}=\dfrac{2}{c}$
so that $c = 2$.
</s>