Prove $\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$ 
Prove $$\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}$$

Hardy uses this fact without proof in a monograph on different ways to evaluate $\int_0^{\infty}\frac{\sin(x)}{x} dx$.
 A: We can use also this well known summation formula (a consequence of the residue theorem): $$\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}f\left(k\right)=-\sum\left\{ \textrm{residues of }\pi\csc\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} 
 $$ so if we take $f\left(z\right)=\frac{1}{x+z\pi}
 $ we get $$\sum_{k\in\mathbb{Z}}\frac{\left(-1\right)^{k}}{x+k\pi}=-\underset{z=-x/\pi}{\textrm{Res}}\left(\frac{\pi\csc\left(\pi z\right)}{x+z\pi}\right)=\csc\left(x\right)$$ as wanted.
A: Presumably the principal value of the two-sided infinite sum is what was intended in the question.
We'll solve this just using Euler's product formula for the sine function:
\begin{align}
\sin x=x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2 \pi^2}\right).
\end{align}
Compute the logarithmic derivative:
\begin{align}
\cot x &= \frac1{x}+\sum_{n=1}^\infty \frac{-2x/n^2\pi^2}{1-\frac{x^2}{n^2 \pi^2}}
\\&=\frac1{x}+\sum_{n=1}^\infty \frac{2x}{x^2-n^2\pi^2}
\\&=\frac1{x}+\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big),
\end{align}
so
$$ \cot x-\frac1{x}=\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big).$$
It follows that
\begin{align}
\cot\big(\frac{x}{2}\big)-\frac2{x}&=\sum_{n=1}^\infty\big(\frac{1}{x/2+n\pi}+\frac{1}{x/2-n\pi}\big)
\\&=2\sum_{n=1}^\infty\big(\frac{1}{x+2n\pi}+\frac{1}{x-2n\pi}\big)
\\&=2\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big).
\end{align}
Subtracting now, we obtain
\begin{align}
\cot\big(\frac{x}{2}\big)-\cot(x)-\frac1{x}&=2\cdot\!\!\!\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)-\sum_{n=1}^\infty\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)
\\&=\sum_{\substack{n\ge 1\\n\text{ even}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)-\sum_{\substack{n\ge 1\\n\text{ odd}}}\big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big)
\\&=\sum_{n=1}^\infty (-1)^n \big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big).
\end{align}
But
\begin{align}
\cot\big(\frac{x}{2}\big)-\cot(x)&=\frac{\cos(x/2)}{\sin(x/2)}-\frac{\cos x}{\sin x}
\\&=\frac{\cos(x/2)}{\sin(x/2)}\cdot\frac{2\sin(x/2)\cos(x/2)}{\sin x}-\frac{\cos x}{\sin x}
\\&=\frac{2\cos^2(x/2)-\cos x}{\sin x}
\\&=\frac1{\sin x}
\\&= \csc x.
\end{align}
So we've shown that
\begin{align}
\csc x &= \frac1{x}+\sum_{n=1}^\infty (-1)^n \big(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\big),
\end{align}
which is the principal value of $$\sum_{n=-\infty}^\infty (-1)^n \frac{1}{x+n\pi},$$
as desired.
A: In this answer (using complex methods) and in this answer (using real methods), it is shown in detail that
$$
\sum_{k=-\infty}^\infty\frac{1}{z+k}=\pi\cot(\pi z)\tag{1}
$$
$(1)$ is the sum for even and odd $k$. The sum for even $k$ would be
$$
\sum_{k=-\infty}^\infty\frac{1}{z+2k}=\frac\pi2\cot\left(\frac{\pi z}2\right)\tag{2}
$$
The sum for even minus the sum for odd would be twice $(2)$ minus $(1)$
$$
\begin{align}
\sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k}
&=\pi\cot\left(\frac{\pi z}2\right)-\pi\cot(\pi z)\\
&=\pi\frac{1+\cos(\pi z)}{\sin(\pi z)}-\pi\frac{\cos(\pi z)}{\sin(\pi z)}\\[9pt]
&=\pi\csc(\pi z)\tag{3}
\end{align}
$$
Therefore,
$$
\sum_{k=-\infty}^\infty\frac{(-1)^k}{z+k\pi}=\csc(z)\tag{4}
$$
A: We begin by expanding the function $\cos(xy)$ in a Fourier series, 
$$\cos(xy)=a_0/2+\sum_{k=1}^\infty a_k\cos(ky) \tag1$$
for $x\in [-\pi/\pi]$.  The Fourier coefficients  $(1)$ are given by
$$\begin{align}
a_k&=\frac{2}{\pi}\int_0^\pi \cos(xy)\cos(ky)\,dy\\\\
&=\frac1\pi (-1)^k \sin(\pi x)\left(\frac{1}{x +k}+\frac{1}{x -k}\right)\tag2
\end{align}$$
Substituting $(2)$ into $(1)$, setting $y=0$, and dividing by $\sin(\pi x)$ reveals
$$\begin{align}
\pi \csc(\pi x)&=\frac1y +\sum_{n=1}^\infty (-1)^k\left(\frac{1}{x -k}+\frac{1}{x +k}\right)\\\\
&=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x-k}\\\\
&=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x+k}\tag3
\end{align}$$
Finally, enforcing the substitution $x\to x/\pi$ and dividing by $\pi$ in $(3)$ yields the coveted result
$$\csc(x)=\sum_{k=-\infty}^\infty \frac{(-1)^k}{x+k\pi}$$
