Show $\frac{\pi}{\alpha \sinh (\pi \alpha)} = \sum_{n= -\infty}^\infty \frac{(-1)^n}{\alpha^2 + n^2} $ The question has 2 parts.
First it asks to show that, given $\;f(t) = e^{\alpha t}\;$ on $\;(-\pi, \pi):$
$$
e^{\alpha t} = \frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) e^{i n t}
$$
which I did using complex version of Fourier expansion:
$$
f(t) = \sum_{n= -\infty}^\infty c_n e^{i n t}
$$
with
$$
 c_n = \frac{1}{2\pi} \int_{-\pi}^\pi f(t) e^{-i n t} dt, \quad \forall\; n \in \mathbb{Z}.
$$
Next it asks using the result from first part, to show that
$$
\frac{\pi}{\alpha \sinh (\pi \alpha)} = \sum_{n= -\infty}^\infty \frac{(-1)^n}{\alpha^2 + n^2}
$$
and that it where I am stuck.
any hints?
 A: You're almost there.  Note that from odd symmetry that 
$$\color{blue}{\sum_{n=-\infty}^\infty \frac{(-1)^n\,n}{n^2+\alpha^2}=\sum_{n=-\infty}^{-1} \frac{(-1)^n\,n}{n^2+\alpha^2}+\sum_{n=1}^\infty \frac{(-1)^n\,n}{n^2+\alpha^2}=0}$$
Now, setting $t=0$ in the equation
$$e^{\alpha t} = \frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) e^{i n t}$$
reveals
$$\begin{align}1 &= \frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) \\\\
&=\frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}\,\alpha}{\alpha^2 + n^2}+i\frac{\sinh (\pi \alpha)}{\pi}\color{blue}{\overbrace{ \sum_{n= -\infty}^\infty \frac{(-1)^{n}\,n}{\alpha^2 + n^2}}^{=0}} \\\\
&=\frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}\,\alpha}{\alpha^2 + n^2}
\end{align}$$
whereupon solving for the series yields
$$\sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}=\frac{\pi}{\alpha \sinh(\pi \alpha)}$$
as was to be shown!
A: Hint: Since $f$ is differentiable at $0$, its Fourier series at $0$ converges to $f(0)$.
Complete answer: You get
\begin{align}
1 &= \frac{\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) \\
&= \frac{\alpha\sinh (\pi \alpha)}{\pi} \sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}\tag{$\star$}
\end{align}
hence
$$
\sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2} = \frac{\pi}{\alpha \sinh(\pi \alpha)}
$$
Note. From the three following facts:


*

*if the series $\displaystyle\sum_{-\infty}^{\infty}$ converges then its value is equal to $\displaystyle\lim_{m\to\infty}\sum_{n=-m}^m$

*the term $a_n:=\frac{(-1)^n}{\alpha^2+n^2}in$ equals $0$ for $n=0$

*$a_n=-a_{-n}$ for $n>0$


we see that
$$
\sum_{n= -\infty}^\infty \frac{(-1)^{n}}{\alpha^2 + n^2}(\alpha + i n) = \alpha \lim_{m\to\infty}\sum_{n= -m}^m\frac{(-1)^{n}}{\alpha^2 + n^2}
$$
But this last series converges absolutely hence the order of summation doesn't matter and indeed we have
$$
\alpha \lim_{m\to\infty}\sum_{n= -m}^m\frac{(-1)^{n}}{\alpha^2 + n^2} = \alpha \sum_{n= -\infty}^{\infty}\frac{(-1)^{n}}{\alpha^2 + n^2}
$$
This justifies $(\star)$.
