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This question already has an answer here:

Calculate $\sum_{n=1}^\infty \frac{1}{n^4}$.

Remark: I know that $\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90}$, but not how to prove that, I totally stalled.

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marked as duplicate by Semiclassical, Winther, Jack D'Aurizio, JKnecht, JMoravitz May 2 '16 at 0:38

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Try using the identity $||f||^2 = \sum_1^\infty |{\hat f}(n)|^2$ where ${\hat f}(n) = \int_0^1 f(x) e^{-2\pi i n x} dx,$ and $f(x) = x^2$ and is periodic with period 1, meaning $f(x) = f(x+1).$ The norm involved would be $||f||^2 = \int_0^1 f(x)^2 dx.$ Do the details out carefully and use the fact that $\sum_1^\infty n^{-2} = \pi^2/6$ to get the result.

Its a quite beautiful way to find the sum, I thought, the first time I worked it out. Do you see how you might change the function $f(x)$ to find $\sum_1^\infty n^{-2} = \pi^2/6$ too?

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