How to prove this identity π=∞∑k=−∞(sin(k)k)2\pi=\sum\limits_{k=-\infty}^{\infty}\left(\frac{\sin(k)}{k}\right)^{2}\;?

Find a function whose Fourier coefficients are $\sin{k}/k$. Then evaluate the integral of the square of that function.

To wit, let

$$f(x) = \begin{cases} \pi & |x|<1\\0&#038;|x|>1 \end{cases}$$

Then, if

$$f(x) = \sum_{k=-\infty}^{\infty} c_k e^{i k x}$$

then

$$c_k = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: f(x) e^{i k x} = \frac{\sin{k}}{k}$$

By Parseval’s Theorem:

$$\sum_{k=-\infty}^{\infty} \frac{\sin^2{k}}{k^2} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: |f(x)|^2 = \frac{1}{2 \pi} \int_{-1}^{1} dx \: \pi^2 = \pi$$

ADDENDUM

This result is easily generalizable to

$$\sum_{k=-\infty}^{\infty} \frac{\sin^2{a k}}{k^2} = \pi\, a$$

where $a \in[0,\pi)$, using the function

$$f(x) = \begin{cases} \pi & |x|<a\\0&#038;|x|>a \end{cases}$$