$\sum_{n=1}^\infty\log\left (\frac{n^2}{1+n^2}\right)$ I'm trying to evaluate the following series:
$$\sum_{n=1}^\infty \log \left(\dfrac{n^2}{1+n^2}\right)$$
- In this case the terms are negative 


*

*$\lim\limits_{n\rightarrow \infty} \log  \left(\dfrac{n^2}{1+n^2}\right)=\log 1=0$

*Now I'm not sure about the application of a test
$\lim\limits_{n\rightarrow \infty} \dfrac {\log \left(\frac{n^2}{1+n^2}\right)}{\left(\frac {1}{2}\right)^n}=0$
being $\sum_{n=1}^\infty \left(\frac {1}{2}\right)^n$ is a geometric series that converges.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{equation}
\mbox{Note that}\quad\sum_{n = 1}^{\infty}\ln\pars{n^{2} \over 1 + n^{2}} =
\ln\pars{\bbox[10px,#ffd]{\ds{%
\lim_{N \to \infty}\prod_{n = 1}^{N}{n^{2} \over n^{2} + 1}}}}
\label{1}\tag{1}
\end{equation}
Then,
\begin{align}
\bbox[10px,#ffd]{\ds{%
\lim_{N \to \infty}\prod_{n = 1}^{N}{n^{2} \over n^{2} + 1}}} & =
\lim_{N \to \infty}\verts{\prod_{n = 1}^{N}{n \over n - \ic}}^{2} =
\lim_{N \to \infty}\verts{N! \over \pars{1 - \ic}^{\large\overline{N}}}^{2}
\\[5mm] & =
\lim_{N \to \infty}\verts{N! \over
\Gamma\pars{1 - \ic + N}/\Gamma\pars{1 - \ic}}^{2} =
\verts{\Gamma\pars{1 - \ic}}^{2}
\lim_{N \to \infty}\verts{N! \over \pars{N - \ic}!}^{2}
\\[5mm] & =
\Gamma\pars{1 - \ic}\
\overbrace{\Gamma\pars{1 + \ic}}^{\ds{\ic\,\Gamma\pars{\ic}}}\
\lim_{N \to \infty}\verts{\root{2\pi}N^{N + 1/2}\expo{-N} \over
\root{2\pi}\pars{N - \ic}^{N - \ic + 1/2}\expo{-\pars{N - \ic}}}^{2}
\\[5mm] & =
\ic\ \overbrace{\quad\bracks{\Gamma\pars{1 - \ic}\Gamma\pars{\ic}}\quad}
^{\ds{{\pi \over \sin\pars{\pi\ic}} = -\ic\,{\pi \over \sinh\pars{\pi}}}}\
\lim_{N \to \infty}\verts{1 \over
N^{-\ic}\pars{1 - \ic/N}^{N - \ic + 1/2}\expo{\ic}}^{2}
\\[5mm] & =
{\pi \over \sinh\pars{\pi}}
\lim_{N \to \infty}\verts{1 \over \expo{-\ic\ln\pars{N}}}^{2} =
\bbox[10px,#ffd]{\ds{{\pi \over \sinh\pars{\pi}}}}\label{2}\tag{2}
\end{align}

\eqref{1} and \eqref{2} lead to

$$
\bbx{\sum_{n = 1}^{\infty}\ln\pars{n^{2} \over 1 + n^{2}} =
\ln\pars{\pi \over \sinh\pars{\pi}}} \approx -1.3018 \\
$$
A: For comparison test with $-\frac{1}{n^2}$
$$\frac{\log \left(\dfrac{n^2}{1+n^2}\right)}{-\frac{1}{n^2}}=\log \left(\dfrac{1+n^2}{n^2}\right)^{n^2}=\log \left(1+\frac1{n^2}\right)^{n^2}\to1$$
A: The given series is convergent by comparison with $\sum_{n\ge1}\frac1{n^2}$. An explicit evaluation can be performed along the following line:
$$\sum_{n\ge1}\log\frac{n^2}{n^2+1}=-\log\prod_{n\ge1}\left(1+\frac1{n^2}\right)=-\log\frac{\sinh\pi}\pi=\color{red}{\log\frac\pi{\sinh\pi}}$$
by invoking $\frac{\sin z}z=\prod_{n\ge1}\left(1-\frac{z^2}{n^2\pi^2}\right)$, i.e. the Weierstrass product for the sine function.
