Computed wrong the sum $\displaystyle\sum_{n\ge1}\frac{\cos n}{n^2} $

Question

I've computed (using standard complex analysis techniques) the sum $$ \sum_{n\ge1}\frac{\cos n}{n^2} $$ and I found $\pi^2/6+1/4$ which is strictly greater than $\pi^2/6$, and this is impossibile, since my series is less or equal than $\sum_{n\ge1}1/n^2$.

So I must have done some mistake, somewhere. I won't write all details of my computation because it would be boring for me and for you too; I only need to know where can I find the sum of this series: in this way I could look at my computations and find the mistake.

Many thanks

Answer 1

If we start from the Fourier series of the sawtooth wave, giving that $$ f(x)=\sum_{n\geq 1}\frac{\sin(nx)}{n}$$ is a $\pi$-periodic function that equals $\frac{\pi-x}{2}$ over $(0,\pi)$, by termwise integration we get that $$ g(x) = \sum_{n\geq 1}\frac{1-\cos(nx)}{n^2} = \frac{\pi x}{2}-\frac{x^2}{4} $$ for any $x\in(0,\pi)$. By evaluating at $x=1$,

$$ \sum_{n\geq 1}\frac{\cos n}{n^2} = \color{red}{\zeta(2)-\frac{\pi}{2}+\frac{1}{4}} $$

follows.

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An alternative approach is given by the Laplace transform (one of my all-time favourites).
Since $$\mathcal{L}^{-1}\left(\frac{1}{x^2}\right)=s, $$ we have: $$ \sum_{n\geq 1}\frac{\cos n}{n^2} = \int_{0}^{+\infty}\sum_{n\geq 0} s\cos(n) e^{-ns}\,ds=-\int_{0}^{+\infty}\frac{s}{2}\left(\frac{1}{e^{s+i}-1}+\frac{1}{e^{s-i}-1}\right)\,ds $$ or: $$ \sum_{n\geq 1}\frac{\cos n}{n}=\int_{0}^{+\infty}\frac{e^s\cos(1)-1}{e^{2s}-2e^{s}\cos(1)+1}\,s\,ds\\ = \int_{0}^{+\infty}\left(s-\log\sqrt{e^{2s}-2e^s\cos(1)+1}\right)\,ds\\ = -\frac{1}{2}\int_{0}^{+\infty}\log\left(1+e^{-2s}-2e^{-s}\cos(1)\right)\,ds\\=-\int_{0}^{1}\frac{\log(1+t^2-2t\cos(1))}{2t}\,dt $$ that can be tackled through the residue theorem or Weierstrass products.
However, I think that the approach through Fourier series is way easier.