Computing $\sum\limits_{k=-\infty}^{+\infty}\frac{\left(-1\right)^{k}}{\alpha k +\beta}$ I wonder how we can prove that
$$
\sum_{k=-\infty}^{+\infty}\frac{\left(-1\right)^{k}}{\alpha k +\beta}=\frac{\pi}{\alpha \sin\left(\displaystyle \frac{\beta}{\alpha}\pi\right)}
$$
without writing it as an integral. I've really no idea on how to proceed, any hints?
 A: We have to make a few assumptions to ensure convergence: the LHS is not absolutely convergent, so it is better to intend $\sum_{k=\infty}^{+\infty}$  in a symmetric fashion, as $\lim_{M\to +\infty}\sum_{k=-M}^{M}$, and $\alpha k+\beta $ has to be non-vanishing over $\mathbb{Z}$, otherwise some term is undefined. With such assumptions,
$$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{\alpha k+\beta} = \frac{1}{\beta}+\frac{2}{\beta}\sum_{k\geq 1}\frac{(-1)^k}{1-\left(\frac{\alpha}{\beta}\right)^2 k^2}.\tag{1}$$
Now it is enough to separate odd/even values of $k$ and to recall that
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{2} $$
$$ \psi(x)-\psi(1-x) = -\pi\cot(\pi x)\tag{3} $$
$$ \frac{1}{\sin x} = \cot\frac{x}{2}-\cot x\tag{4} $$
to recover the wanted identity.
A: Assume we may write some function $f$ as a product over its roots, i.e. we may write
$$f(x)=\prod_{f(w)=0}(x-w).$$
Then on one hand we see that 
$$\frac1{f(x)}=\prod_{f(w)=0}\frac1{x-w}$$
and we can do some fraction decomposition and assume that we may write 
$$\prod_{f(w)=0}\frac1{x-w}=\sum_{f(w)=0}\frac{b(w)}{x-w}$$
for some coefficients $b(w)$. We can multiply through by $f(x)$ and see that this becomes
$$1=\sum_{f(w)=0}b(w)\prod_{{f(a)=0}\atop{a\ne w}}(x-a)$$
so we plug in $x=\phi$ for some $f(\phi)=0$ and see that every term on the RHS vanishes except for the case $w=\phi$ and we're left with
$$b(\phi)=\prod_{{f(a)=0}\atop{a\ne \phi}}\frac1{\phi-a}.$$
On the other hand, we have 
$$\ln f(x)=\sum_{f(w)=0}\ln(x-w)$$
which implies that 
$$\frac{f'(x)}{f(x)}=\sum_{f(w)=0}\frac1{x-w}$$
which is
$$f'(x)=\sum_{f(w)=0}\prod_{{f(a)=0}\atop{a\ne w}}(x-a).$$
Then assuming that $f(w)=0\Rightarrow f'(w)\ne 0$ we have, for any $f(\phi)=0$
$$f'(\phi)=\prod_{{f(a)=0}\atop{a\ne\phi}}(\phi-a)$$
and consequently 
$$\frac1{f(x)}=\sum_{f(w)=0}\frac1{(x-w)f'(w)}.\tag 1$$
We take this a step further by seeing that the roots of $f(x)=\sin\pi x$ are the integers. Indeed, it was Euler who wrote that
$$\sin\pi x=\pi x\prod_{n\ge1}\left(1-\frac{x^2}{n^2}\right)=\pi x\prod_{n\ne0}\left(1+\frac{x}{n}\right).$$
We apply $(1)$ to $f(x)=\sin\pi x$ and get
$$\frac1{\sin\pi x}=\sum_{k\in\Bbb Z}\frac{1}{(x+k)\pi\cos\pi k}=\frac1\pi\sum_{k\in\Bbb Z}\frac{(-1)^k}{x+k}.$$
Plugging in $x=\beta/\alpha$ gives the result in question.
