Series of $\csc(x)$ or $(\sin(x))^{-1}$

Question

In some cases I found that $$\csc(x)= \lim\limits_{k\rightarrow \infty}\sum_{n=-k}^{k}(-1)^{n}\frac{1}{x-n\pi}$$

Is anything to prove or disprove that?

Answer 1

Recall that the sine function can be represented as the infinite product

$$\sin x=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right) \tag 1$$

Taking the logarithmic derivative of $(1)$, we obtian

$$\begin{align} \cot x&=\frac1x+2x\sum_{n=1}^{\infty}\frac{1}{x^2-n^2\pi^2}\\\\ &=\sum_{n=-\infty}^{\infty}\frac{x}{x^2-n^2\pi^2}\\\\ &=\frac12\sum_{n=-\infty}^{\infty}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right) \end{align}$$

Then, using the trigonometric identity $\csc x=\cot(x/2)-\cot x$ reveals that

$$\begin{align} \csc x&=\frac12\sum_{n=-\infty}^{\infty}\left(\frac{2}{x-2n\pi}-\frac{1}{x-n\pi}+\frac{2}{x+2n\pi}-\frac{1}{x+n\pi}\right)\\\\ &=\sum_{n=-\infty}^{\infty}\left(\frac{2}{x-2n\pi}-\frac{1}{x-n\pi}\right)\\\\ &=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{x-n\pi} \end{align}$$

as was to be shown!