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

• use residue theorem ... – Math-fun Dec 16 '14 at 8:40
• A related question. – Lucian Dec 16 '14 at 8:41
• @Mehdi Just use math analysis... – xldd Dec 16 '14 at 9:27

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