I'mma $f(n)$ Maths You Up: 0

I was reading MathOverflow and a certain comment caught my eye. Sadly, I can't (be bothered) find(ing) the link to the actual comment, so I'll just have to furnish you with following attempt at recreating the comment:

"Yo, dawg. I knows you be trippin', coz I too be trippin' the first time I saw da followin' truth: $$\begin{align*} \sum_{k=1}^{\infty}\frac{\cos(kx)}{k^2}=\frac{1}{2}(\mathrm{Li}_{2}\exp(ix)+\mathrm{Li}_{2}\exp(-ix)) \end{align*}$$ be a pe-ri-o-dik para-bo-la. And you can bank dat in yo' mama's thesis."

I may have exercised some liberties in choosing the vernacular for this non-verbatim quote.

In any case, my first thought was: NO $f(n)$ WAY.

And then I thought, wait - what does he mean by a periodic parabola? So I plotted the function in (were)Wolfram Alpha, and sure enough...

...it's a bleeding periodic parabola.

And when you think a bit more about it, maybe it shouldn't have been THAT surprising to me. I mean, the derivative of this series is really just: $$\begin{align*} \sum_{k=1}^{\infty}\frac{\sin(kx)}{k} \end{align*}$$ Which is a famous example of a Fourier series called a sawtooth wave. So it's periodic and each period is a linear function. Which means that its integral, that is: our original series, must be a periodic parabola.

Still, that's $f(n)$ mathed up. =)