<s>[INST] 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 $\int_0^{\infty}\frac{\sin(x)}{x} dx$.
 [/INST] 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.
</s>