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I need to find the exact value of the power series $-\sum_{k=1}^{\infty}\frac{(-1)^k}{1+k^2}$

It looks like it could have something to do with $\arctan x$ or some sort of trigonometric function, but I don't have a clue where to start.

It would be great if there are any hints on how to proceed. Thank you!

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    $\begingroup$ Consider the function $$f(z) = \frac{\pi}{\left(1+z^2\right)\sin(\pi z)}$$ The contour integral along a circle of radius $R$ tends to zero if $R$ is sent to infinity. If you express the contour integral as the sum of all the residues at the poles inside the contour, then you get your summation (also over negative $k$ but summand is even), and there are two poles at $z = \pm i$. So, the summation can be expressed in terms of the residues at $z = \pm i$. $\endgroup$ – Count Iblis Feb 23 '18 at 2:02
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Avoiding Complex Analysis, a simple way is to notice that the given series is absolutely convergent and it is related to the Fourier (cosine) series of $e^{-|x|}$ over $(-\pi,\pi)$. Indeed $$ \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-|x|}\,dx = \frac{1-e^{-\pi}}{\pi},\qquad \frac{1}{\pi}\int_{-\pi}^{\pi}e^{-|x|}\cos(nx)\,dx = \frac{2}{\pi(n^2+1)}\left(1-(-1)^n e^{-\pi}\right) $$ imply $$ e^{-|x|}\stackrel{L^2}{=}\frac{1-e^{-\pi}}{\pi}+\frac{2}{\pi}\sum_{n\geq 1}\frac{\cos(nx)(1-(-1)^n e^{-\pi})}{n^2+1} $$ but we also have pointwise convergence over $[-\pi,\pi]$ due to the summability of $\frac{1}{n^2+1}$. In particular we are allowed to just evaluate both sides at $x=\pi$ and $x=0$ to get: $$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{k^2+1} = \frac{1}{2}\left(1-\frac{\pi}{\sinh \pi}\right).$$

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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} -\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over 1 + k^{2}} & = \Im\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k + \ic} = {1 \over 2}\,\Im\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 + \ic/2} - {1 \over k + 1/2 + \ic/2}} \\[5mm] & = {1 \over 2}\,\Im\bracks{\Psi\pars{{1 \over 2} + {1 \over 2}\,\ic} - \Psi\pars{1 + {1 \over 2}\,\ic}}\qquad\pars{~\Psi:\ Digamma\ Function~} \end{align}

See $\color{#000}{\mathbf{6.3.12}}$ and $\color{#000}{\mathbf{6.3.13}}$, respectively, in A&S Table:

$\left\{\begin{array}{rcl} \ds{\Im\Psi\pars{{1 \over 2} + {1 \over 2}\,\ic}} & \ds{=} & \ds{\phantom{-}{1 \over 2}\,\pi\tanh\pars{\pi \over 2}} \\[1mm] \ds{\Im\Psi\pars{1 + {1 \over 2}\,\ic}} & \ds{=} & \ds{-1 + {1 \over 2}\,\pi\coth\pars{\pi \over 2}} \end{array}\right.$

such that $$ \bbx{-\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over 1 + k^{2}} = {1 \over 2}\bracks{1 - \pi\,\mrm{csch}\pars{\pi}} \approx 0.3640} $$

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