Series sum $\sum 1/(n^2+(n+1)^2)$ In an exercise, I caculate the Fourier expansion of $e^x$ over $[0,\pi]$ is
$$e^x\sim \frac{e^\pi-1}{\pi}+\frac{2(e^\pi-1)}{\pi}\sum_{n=1}^\infty \frac{\cos 2nx}{4n^2+1}+\frac{4(1-e^\pi)}{\pi}\sum_{n=1}^\infty \frac{n\sin 2nx}{4n^2+1}.$$
From this, it is easy to deduce 
$$\sum_{n=1}^\infty \frac{1}{4n^2+1}=\frac{\pi}{4}\frac{e^\pi+1}{e^\pi-1}-\frac{1}{2}.$$
However, I could not find the following sum
$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2+1},$$
from which we can calculate the sum $\sum 1/(n^2+1)$.
 A: We can approach such kind of series by considering logarithmic derivatives of Weierstrass products. For instance, from:
$$\cosh z = \prod_{n=1}^{+\infty}\left(1+\frac{4z^2}{(2n-1)^2\pi^2}\right)\tag{1}$$
we get:
$$\frac{\pi}{2}\tanh\frac{\pi z}{2} = \sum_{n=1}^{+\infty}\frac{2z}{z^2+(2n-1)^2}\tag{2},$$
so, evaluating in $z=1$:
$$\sum_{n=1}^{+\infty}\frac{1}{(2n-1)^2+1}=\color{red}{\frac{\pi}{4}\tanh\frac{\pi}{2}}.\tag{3}$$
With the same approach, but starting from the Weierstrass product for $\frac{\sinh z}{z}$, we can compute $\sum_{n\geq 1}\frac{1}{1+n^2}$, too:
$$\sum_{n\geq 1}\frac{1}{n^2+1}=\frac{-1+\pi\coth\pi}{2}.\tag{4}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1}^{2} + 1}}
=\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1 + \ic}\pars{2n - 1 - \ic}}
\\[5mm]&=\sum_{n\ =\ 0}^{\infty}{1 \over \pars{2n + 1 + \ic}\pars{2n + 1 - \ic}}
={1 \over 4}\sum_{n\ =\ 0}^{\infty}
{1 \over \bracks{n + \pars{1 + \ic}/2}\bracks{n + \pars{1 - \ic}/2}}
\\[5mm]&={1 \over 4}\,
{\Psi\pars{\bracks{1 + \ic}/2} - \Psi\pars{\bracks{1 - \ic}/2}\over
 \pars{1 + \ic}/2 - \pars{1 - \ic}/2}
\end{align}
where $\ds{\Psi}$ is the Digamma Function.

Then,
  $$
\color{#66f}{\large%
\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1}^{2} + 1}}
={1 \over 4}\,\ic\bracks{\Psi\pars{1 - \ic \over 2} - \Psi\pars{1 + \ic \over 2}}
$$

With Euler Reflection Formula
$\ds{\Psi\pars{1 - x} - \Psi\pars{x} = \pi\cot\pars{\pi x}}$:
\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 1}^{\infty}{1 \over \pars{2n - 1}^{2} + 1}}
={1 \over 4}\,\ic\ \pi\cot\pars{\pi\,{1 + \ic \over 2}}
=-\,{1 \over 4}\,\ic\,\pi\tan\pars{\pi\ic \over 2}
\\[5mm]&=-\,{1 \over 4}\,\ic\,\pi\bracks{\ic\tanh\pars{\pi \over 2}}
=\color{#66f}{\large{1 \over 4}\,\pi\tanh\pars{\pi \over 2}}
\end{align}
