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 $\sum 1/(n^2+1)$.

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:
$$\sum_{n\geq 1}\frac{1}{n^2+1}=\frac{-1+\pi\coth\pi}{2}.\tag{4}$$
