# 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.

• According to Mathematica, $$\sum_{n=1}^{\infty } \log \left(1+\frac{1}{n^2}\right) = i \pi +\log (2)-\log (-\Gamma (2-i))-\log (\Gamma (2+i))$$ – Math1000 Sep 10 '16 at 7:43

## 3 Answers

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $$\Gamma\pars{1} = 1\,,\qquad \Gamma\pars{z + 1} = z\,\Gamma\pars{z}\,,\qquad \Gamma\pars{z}\Gamma\pars{1 - z} = {\pi \over \sin\pars{\pi z}}$$

• you should forget about $\Gamma$ and explain how we know the partial fraction expansion of $\ln(\sinh(z)/\pi)$ (so it would be a proof of the Weierstrass product for $\sin(z)/z$) – reuns Sep 10 '16 at 3:07
• @user1952009 I'm not using such partial fraction expansion. – Felix Marin Sep 10 '16 at 3:10
• I meant the partial fraction expansion of $\frac{\cosh(z)}{\sinh(z)}$ so that you know the $\sum_n \ln(1+ z^2 / n)$ series for $\ln( \sinh(z)/\pi)$ (and of course you are) – reuns Sep 10 '16 at 3:14
• @user1952009 Other answer already exploits that point of view. So, I'm forced to write something different. By the way, such expansions are derived via Mittag-Leffler expansion, Residue Theorem and so on... – Felix Marin Sep 10 '16 at 3:25

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.

• The asker of the question only needed a hint. Moreover, no context or original effort is given in the question. Just because you have enough mathematical knowledge to be able to furnish detailed answers, does not mean you should. – heropup Sep 9 '16 at 23:16
• @heropup: But an answer is not only for the question OP, its purpose is to serve the whole community. What should we do, perform a test to check if the asker deserves a good answer, a mediocre answer or no answer? – Jack D'Aurizio Sep 10 '16 at 0:27

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.