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

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.

• Since $\log(1+x)\sim x$ as $x\to 0$ we have $\log(1 - \frac{1}{1+n^2}) \sim - \frac{1}{1+n^2}$. If you want to use the comparison test you should try to compare to the series $\sum \frac{1}{1+n^2}$ (or simply $\sum \frac{1}{n^2}$). – Winther Jan 17 '18 at 17:14
• $\sum _{n=1}^{\infty } \log \left(\frac{n^2}{1+n^2}\right)=\log (\pi \text{csch}(\pi ))$ – Mariusz Iwaniuk Jan 17 '18 at 17:21
• @MariuszIwaniuk Sure, but showing the OP how to derive or prove this is sort of the crux of the matter. – Clement C. Jan 17 '18 at 17:22
• @Mariusz Iwaniuk You should make that an answer and show some detail, it would be interesting for those who don't know those things, yet. – Professor Vector Jan 17 '18 at 17:23
• Anne: to prove convergence, @Winther's approach is one of the simplest (I would actually recommend writing $-\log\frac{1+n^2}{n^2} = -\log\left(1+\frac{1}{n^2}\right) \sim_{n\to\infty} -\frac{1}{n^2}$ to avoid pesky $+1$'s). Now, why did you decide to compare it to a geometric series -- what prompted you to do so? (it doesn't work, so understand why you chose to try that is crucial to helping you). – Clement C. Jan 17 '18 at 17:25

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

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

• But besides showing convergence, the OP's clearly stated goal is to evaluate the sum. – Clement C. Jan 17 '18 at 17:38
• @ClementC. Anne has declared her doubt on the test, I've asked a clarification on the OP. Thanks! – user Jan 17 '18 at 17:40
• @ClementC $\lim_{x\rightarrow 0} \frac{log(1+x)}{x}=\lim_{x\rightarrow 0} \frac{1}{1+x}=1$ and then $log (1+x) \sim x$ for $x \rightarrow 0$ $log(\frac{n^2}{1+n^2})=log(\frac{1}{\frac{1+n^2}{n^2}})=-log (1+ \frac {1}{n^2}) \sim - \frac {1}{n^2}$ for $n \rightarrow \infty$ – Anne Jan 18 '18 at 14:12
• @ClementC. Applying the asindotic comparison test: $\frac{log(\frac{n^2}{1+n^2} )}{- \frac {1}{n^2}}=\frac{-log(\frac{n^2}{1+n^2} )}{- \frac {1}{n^2}}=\frac{log(\frac{n^2}{1+n^2} )}{\frac {1}{n^2}}\rightarrow 1$ for $n \rightarrow \infty$. Since $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, $\sum_{n=1}^\infty log(\frac{n^2}{1+n^2})$ converges, too. is it correct? – Anne Jan 18 '18 at 14:12
• @Anne Yes Anne it is correct! The comparison test is valid for series whiche terms are eventually positive or negative. – user Jan 18 '18 at 14:25

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.

• I want to learn more about these things, in a systematic manner. please tell me path. just out of high school. Thank for good answer – King Tut Jan 17 '18 at 18:24
• @KingTut: have a look at my notes, I wish you may learn some intersting tricks there. – Jack D'Aurizio Jan 17 '18 at 18:25
• @JackD'Aurizio Thanks for the reference, I also will have a look to it! – user Jan 18 '18 at 14:30

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$\mbox{Note that}\quad\sum_{n = 1}^{\infty}\ln\pars{n^{2} \over 1 + n^{2}} = \ln\pars{\bbox[10px,#ffd]{\ds{% \lim_{N \to \infty}\prod_{n = 1}^{N}{n^{2} \over n^{2} + 1}}}} \label{1}\tag{1}$$ 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}

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