Prove that $\prod_{k=2}^{+\infty} (1+1/k^2) = \sinh(\pi)/(2 \pi)$. My attempt 1: Let $x_n=\left(1+\frac{1}{2^2} + \cdots + \frac{1}{n^2} \right)$, and we have $x_{n+1}>x_n$. Since
$$
1+\frac{1}{n^2} \le 1+ \frac{1}{(n-1)(n+1)} = \frac{n}{n-1} \cdot \frac{n}{n+1},
$$
then $ x_n < \frac{2n}{n+1} <2$. Hence $\{x_n\}_{n=2}^{\infty}$ converges. Let
$$
\lim_{n \to \infty} x_n =a,
$$
and notice that $x_{n+1}=\left(1+\frac{1}{(n+1)^2}\right)x_n $. By this, we can only get $a=a$. We can't know the value of $a$.
My attempt 2: We write
$$
\prod_{k=2}^n \left(1+\frac{1}{k^2} \right)=\exp\left(\sum_{k=2}^n \log\left(1+\frac{1}{k^2}\right)\right)
$$
So it suffices to know what's the limit of the serise on the right. I still don't know how to finish it.
Finally,I used Mathematica to calculate the limit, and it tells me that is $\frac{\sinh(\pi)}{2\pi}$. But I don't know how to know it without computer. Can you help me?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{N \in \mathbb{N}_{\geq 2}}$:

\begin{align}
&\bbox[10px,#ffd]{\prod_{k = 2}^{N}\pars{1 + {1 \over k^{2}}}} =
{\bracks{\prod_{k = 2}^{N}\pars{k + \ic}}
\bracks{\prod_{k = 2}^{N}\pars{k - \ic}} \over
\bracks{\prod_{k = 2}^{N}k}\bracks{\prod_{k = 2}^{N}k}}
\\[5mm] = &\
{\verts{\prod_{k = 2}^{N}\pars{k + \ic}}^{2} \over \pars{N!}^{2}}
=
{\verts{\pars{2 + \ic}^{\overline{N - 1}}}^{2} \over \pars{N!}^{2}}
\\[5mm] = &\
{\verts{\Gamma\pars{2 + \ic + N - 1}/\Gamma\pars{2 + \ic}}^{2} \over \pars{N!}^{2}}
\\[5mm] = &\
{1 \over \Gamma\pars{2 + \ic}\Gamma\pars{2 - \ic}}\,
\verts{\pars{N + \ic}! \over N!}^{2}
\\[1cm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim} &\
{1 \over \pars{1 + \ic}\Gamma\pars{1 + \ic}
\pars{1 - \ic}\Gamma\pars{1 - \ic}}\times
\\[2mm] &\
\verts{\root{2\pi}\pars{N + \ic}^{N + \ic + 1/2}\expo{-N - \ic} \over \root{2\pi}N^{N + 1/2}\expo{-N}}^{2}
\\[1cm] = &\
{1 \over 2\,\ic\,\Gamma\pars{\ic}
\Gamma\pars{1 - \ic}}\,
\verts{N^{N + \ic + 1/2}\pars{1 + \ic/N}^{N} \over
N^{N + 1/2}}^{2}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\to} &\
{1 \over 2\ic}\,{1 \over \pi/\bracks{\sin\pars{\pi\ic}}}
=
{1 \over 2\ic}\,{\ic\sinh\pars{\pi} \over \pi} =
\bbx{\sinh\pars{\pi} \over 2\pi}
\end{align}
