$\sum_{n=1}^\infty\log\left (\frac{n^2}{1+n^2}\right)$

Question

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.

Answer 1

For comparison test with $-\frac{1}{n^2}$

$$\frac{\log \left(\dfrac{n^2}{1+n^2}\right)}{-\frac{1}{n^2}}=\log \left(\dfrac{1+n^2}{n^2}\right)^{n^2}=\log \left(1+\frac1{n^2}\right)^{n^2}\to1$$

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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

$$ \bbx{\sum_{n = 1}^{\infty}\ln\pars{n^{2} \over 1 + n^{2}} = \ln\pars{\pi \over \sinh\pars{\pi}}} \approx -1.3018 \\ $$