How do I compute this limit: $\sum_{k = 1}^{\infty} \frac{\sin k}{k}$?

Question

My task is to compute the limit of $$\sum_{k = 1}^{\infty} \dfrac{\sin k}{k}$$, with Fourier-theory.

The only thing I know is that $$\dfrac{\sin k}{k}$$ are the coefficients of the Fourier series $$\pi^{-1} + \dfrac{2}{\pi} \cdot \sum_{k=1}^{n} \dfrac{\sin k}{k} \cos(kx),$$ which builds a square-function.

But I have no idea how to compute the limit with that knowledge and the knowledge about Fourier theory in general.

Answer 1

The Fourier series of $f(x)=\frac{\pi-x}{2}$ for $x\in(0,2\pi)$ and $f(0)=0$ is $\tilde f(x)=\sum_{k=1}^{\infty}\frac{sin(kx)}{k}$.
Now let $x=1$ and you have $\sum_{k=1}^{\infty}\frac{sin(k)}{k}=\frac{\pi-1}{2}$