Series of $-\sum_{k=1}^{\infty}\frac{(-1)^k}{1+k^2}$ 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!
 A: 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:
$$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{k^2+1} = \frac{1}{2}\left(1-\frac{\pi}{\sinh \pi}\right).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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
$$
\bbx{-\sum_{k = 1}^{\infty}{\pars{-1}^{k} \over 1 + k^{2}} =
{1 \over 2}\bracks{1 - \pi\,\mrm{csch}\pars{\pi}} \approx 0.3640}
$$
