"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. =)
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