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The question:

Calculate the sum $$I:=\sum_{n=1}^\infty {(-1)^n\over 1+n^2}$$

My attempt:

  • Notation: In a previous question I have calculated $$\sum_{n=1}^\infty{1\over n^2+1}={1\over 2}\left(\pi{e^\pi+e^{-\pi}\over e^\pi-e^{-\pi}}-1\right)$$ and if possible, I would like to use it.

On the one hand: $$\sum_{-\infty}^\infty {(-1)^n\over 1+n^2}=1+2I$$ On the other hand:

$$\sum_{-\infty}^\infty {(-1)^n\over 1+n^2}=-Res((-1)^z\cdot{\pi \cot(\pi z)\over 1+z^2},i)-Res((-1)^z\cdot{\pi\cot(\pi z)\over 1+z^2},-i) \\ Res((-1)^z\cdot{\pi \cot(z\pi)\over 1+z^2},i)={(-1)^i\over 2i}\cdot\cot(\pi i) \\ Res((-1)^z\cdot{\pi \cot(z\pi)\over 1+z^2},-i)={(-1)^{-i}\over -2i}\cdot\cot(-\pi i) $$ And in general: $$ \sum_{-\infty}^\infty={\pi\over 2i}((-1)^i\cot(-\pi i)-(-1)^i\cot(\pi i)) $$ But I don't know how to keep evaluate it.

Bernard
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J. Doe
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  • It is $$\frac{1}{2} (\pi \text{csch}(\pi )-1)$$ – Dr. Sonnhard Graubner May 27 '19 at 17:34
  • Sorry, can you please share the question where you evaluated $$\sum_{n=1}^\infty{1\over n^2+1}={1\over 2}\left(\pi{e^\pi+e^{-\pi}\over e^\pi-e^{-\pi}}-1\right)$$? – BowPark May 28 '19 at 07:44
  • There's a theorem: Let $f:\mathbb{C}\setminus {a_1,\dots a_n}\to\mathbb{C}$ such that $f$ doesn't have singularities in $\mathbb{Z}$ and for all $|z|>R$ (for some $R\in\mathbb{R}$), $|f(z)|<{c\over |z|^{1+\epsilon}}$. Then $$\sum_{-\infty}^\infty f(n)=-\sum_i Res(f\cdot \phi,a_i)$$ where $a_i$ are the singularities of $f$ and $$\phi(z)=\pi\cot(\pi z)$$ @BowPark – J. Doe May 28 '19 at 15:09

3 Answers3

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If you know how to compute $\sum_{n\geq 1}\frac{1}{n^2+a^2}$ (for instance, through the Poisson summation formula) then you may just exploit $$ \sum_{n\geq 1}\frac{(-1)^n}{n^2+1}=-\sum_{n\geq 1}\frac{1}{n^2+1}+2\sum_{m\geq 1}\frac{1}{(2m)^2+1}=\frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2+\left(\frac{1}{2}\right)^2}-\sum_{n\geq 1}\frac{1}{n^2+(1)^2}.$$ Since $\int_{0}^{+\infty}\frac{\sin(nx)}{n}e^{-ax}\,dx = \frac{1}{n^2+a^2}$ we also have $$ \sum_{n\geq 1}\frac{1}{n^2+a^2} = \int_{0}^{+\infty}W(x) e^{-ax}\,dx = \frac{1}{1-e^{-2\pi a}}\int_{0}^{2\pi}\frac{\pi -x}{2}e^{-ax}\,dx$$ where $W(x)$ is $2\pi$-periodic and piecewise-linear sawtooth wave, which equals $\frac{\pi-x}{2}$ over $(0,2\pi)$.
This immediately leads to $$ \sum_{n\geq 1}\frac{(-1)^n}{n^2+1} = \frac{\pi}{e^{\pi}-e^{-\pi}}-\frac{1}{2}.$$

Jack D'Aurizio
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$$\sum_{n=1}^\infty {(-1)^n\over 1+n^2}=\sum_{n=1}^\infty {1\over 1+(2n)^2}-\sum_{n=1}^\infty {1\over 1+(2n-1)^2}$$ we know that $$\frac{\pi x\coth(\pi x)-1}{2x^2}=\sum_{n=1}^{\infty}\frac{1}{x^2+n^2}=\frac{1}{x^2}\sum_{n=1}^{\infty}\frac{1}{1+(\frac{n}{x})^2}$$ $$\frac{\pi x\coth(\pi x)-1}{2}=\sum_{n=1}^{\infty}\frac{1}{1+(\frac{n}{x})^2}$$

let$x=\frac{1}{2}$ $$\frac{\frac{\pi}{2}\coth(\frac{\pi}{2})-1}{2}=\sum_{n=1}^{\infty}\frac{1}{1+(2n)^2}\tag1$$ and we have $$\frac{\pi \tanh(\pi x/2)}{4x}=\sum_{n=1}^{\infty}\frac{1}{x^2+(2n-1)^2}$$ let $x=1$ $$\frac{\pi \tanh(\pi /2)}{4}=\sum_{n=1}^{\infty}\frac{1}{1+(2n-1)^2}\tag2$$ so $$\sum_{n=1}^\infty {(-1)^n\over 1+n^2}=\frac{\frac{\pi}{2}\coth(\frac{\pi}{2})-1}{2}-\frac{\pi \tanh(\pi /2)}{4}$$

E.H.E
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2} + 1}} = \Im\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n - \ic} = \Im\sum_{n = 0}^{\infty} \pars{{1 \over 2n + 2 - \ic} - {1 \over 2n + 1 - \ic}} \\[5mm] = &\ {1 \over 2}\,\Im\sum_{n = 0}^{\infty} \pars{{1 \over n + 1 - \ic/2} - {1 \over n + 1/2 - \ic/2}} = {1 \over 2}\,\Im\bracks{\Psi\pars{{1 \over 2} - {\ic \over 2}} - \Psi\pars{1 - {\ic \over 2}}} \end{align} where $\ds{\Psi}$ is the Digamma Function. See $\ds{\mathbf{\color{black}{6.3.16}}}$ in A & S Table

Then, \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2} + 1}} = {1 \over 2}\braces{{\bracks{\Psi\pars{1/2 - \ic/2} - \Psi\pars{1/2 + \ic/2}} \over 2\ic} - {\bracks{\Psi\pars{1 - \ic/2} - \Psi\pars{1 + \ic/2}} \over 2\ic}} \end{align} With $\ds{\Psi}$-Recurrence and Euler Reflection Formula: \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2} + 1}} = -\,{1 \over 4}\,\ic\braces{% \pi\cot\pars{\pi\bracks{{1 \over 2} + {\ic \over 2}}} - \pi\cot\pars{\pi\,{\ic \over 2}} + {1 \over \ic/2}} \\[5mm] = & \bbx{\pi\,\mrm{csch}\pars{\pi} - 1 \over 2} \approx -0.3640 \end{align}
Felix Marin
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