What function does the Fourier series $\pi^2 / 6 + \sum^{\infty}_{k=1} \frac {-1} {k^2} \cos(kx) $ converge against?

Question

What function does the Fourier series $$\pi^2 / 6 + \sum^{\infty}_{k=1} \frac {-1} {k^2} \cos(kx) = \pi^2 / 6 + \sum^{\infty}_{k=1} \frac {-1} {2k^2} (e^{ikx} + e^{-ikx})$$ converge against ?

I've proved that the Fourier series converge uniformly using Weierstrass M-test. However, I don't see what function is converge against ?

Answer 1

Hint: Consider $k\in\mathbf Z\setminus\{0\}$. $$ \frac{1}{2\pi} \int_0^{2\pi} \frac{x^2-2\pi x}{2} \,\cos(k x)\,\mathrm dx=\frac{1}{k^2}.$$

Answer 2

I think it may be the function $f(x) =\frac{-x^2}{4} +\frac{\pi x }{2}$ for $0<x<2\pi$