Prove: $$\sum_{n=1}^{\infty} \frac{1}{n^2}\cos(n x)= \frac{\pi^2}{6}-\frac{\pi x}{2}+\frac{x^2}{4}$$ for $0\le x\le 2 \pi$
I've found the result in 1.443 I.S. Gradshteyn and I.M. Ryzhik's book. But I need the proof.
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How much do you know about Fourier series? – Umberto P. Apr 11 '17 at 21:12
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@UmbertoP. Ok, maybe I know the proof. – 346699 Apr 11 '17 at 21:15
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@UmbertoP. But the original question is to calculate the infinite sum. What I do now is to find the result in some book and then prove it. So is there some way to calculate the infinite sum directly? – 346699 Apr 11 '17 at 21:18
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1It is pretty well-known and not difficult to prove that the following series $$ \sum_{n\geq 1}\frac{\sin(nx)}{n} $$ is pointwise convergent to a $\pi$-periodic function that over the interval $(0,2\pi)$ equals $\frac{\pi-x}{2}$. By performing a termise integration, we get that $$ \sum_{n\geq 1}\frac{1-\cos(nx)}{n^2} = \frac{\pi^2}{6}-\sum_{n\geq 1}\frac{\cos(nx)}{n^2}$$ is uniformly convergent to the graph of a parabola on the interval $[0,2\pi]$, precisely $\frac{2\pi x-x^2}{4}$. – Jack D'Aurizio Apr 11 '17 at 21:20