Find the sum of series $\sum_{n=1}^\infty \ln(1+1/n^2)$ Is it possible to find the sum of the series $\sum_{n=1}^{\infty}\ln(1+1/n^2)$? Any hint will be appreciated.
 A: We have the Weierstrass product for the sine function:
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right)\tag{1} $$
whose consequence is:
$$ \frac{\sinh(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1+\frac{x^2}{n^2}\right)\tag{2} $$
By evaluating both sides of $(2)$ at $x=1$ and switching to logarithms:
$$ \sum_{n\geq 1}\log\left(1+\frac{1}{n^2}\right) = \color{red}{\log\sinh\pi-\log\pi}\tag{3}$$
follows.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty}\ln\pars{1 + {1 \over n^{2}}}} & =
\sum_{n = 1}^{\infty}\int_{0}^{1}{2x \over x^{2} + n^{2}}\,\dd x =
-\ic\int_{0}^{1}\sum_{n = 0}^{\infty}
\pars{{1 \over n + 1 - x\ic} - {1 \over n + 1 + x\ic}}\,\dd x
\\[5mm] & =
-\ic\int_{0}^{1}
\bracks{\Psi\pars{1 + x\ic} - \Psi\pars{1 - x\ic}}\,\dd x\qquad
\pars{~\Psi:\ Digamma\ Function~}
\end{align}

Since
$\ds{\Psi\pars{z} \stackrel{\mrm{def.}}{=}
\totald{\ln\pars{\Gamma\pars{z}}}{z}\,,\quad\pars{~\Gamma:\ Gamma\ Function~}}$:
\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty}\ln\pars{1 + {1 \over n^{2}}}} & =
-\ln\pars{\Gamma\pars{1 + \ic}} - \ln\pars{\Gamma\pars{1 - \ic}} =
-\ln\pars{\ic\,\Gamma\pars{\ic}\Gamma\pars{1 - \ic}}
\\[5mm] & =
-\ln\pars{\ic\,{\pi \over \sin\pars{\pi\ic}}} =
-\ln\pars{\ic\,{\pi \over \sinh\pars{\pi}\ic}} =
\color{#f00}{\ln\pars{\sinh\pars{\pi} \over \pi}}
\end{align}


Note that
  $$
\Gamma\pars{1} = 1\,,\qquad
\Gamma\pars{z + 1} = z\,\Gamma\pars{z}\,,\qquad
\Gamma\pars{z}\Gamma\pars{1 - z} = {\pi \over \sin\pars{\pi z}}
$$

A: We have $$\prod_{n\geq1}\left(1+\frac{1}{n^{2}}\right)=\prod_{n\geq1}\frac{n^{2}+1}{n^{2}}=\prod_{n\geq0}\frac{\left(n+\left(1-i\right)\right)\left(n+\left(1+i\right)\right)}{\left(n+1\right)\left(n+1\right)}
 $$ and now we can use the identity $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d
 $$ so $$\prod_{n\geq1}\left(1+\frac{1}{n^{2}}\right)=\frac{1}{\Gamma\left(1-i\right)\Gamma\left(1+i\right)}
 $$ and so $$\sum_{n\geq1}\log\left(1+\frac{1}{n^{2}}\right)=-\log\left(\Gamma\left(1-i\right)\Gamma\left(1+i\right)\right)=\color{red}{\log\left(\frac{\sinh\left(\pi\right)}{\pi}\right)}$$ from the reflection formula of the Gamma function.
