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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?

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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.

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  • $\begingroup$ I don't think it must be interpreted as a symmetric limit of the upper and lower bounds. Pretty sure $\lim\limits_{a,b\to\infty}\sum_{k=-a}^b$ converges. $\endgroup$ – Simply Beautiful Art Feb 3 '18 at 15:13
  • $\begingroup$ @SimplyBeautifulArt: correct, re-worded. $\endgroup$ – Jack D'Aurizio Feb 3 '18 at 15:16
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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.

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