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Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series.

What function do I pick for the summation from $$\sum_{n =1}^{\infty}\frac{1}{n^6} \ ?$$ using Parseval's identity

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marked as duplicate by J. M., Rasmus, Asaf Karagila, Beni Bogosel, Zev Chonoles Apr 27 '12 at 20:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

i dont know what that is? – Hala Shubair Apr 27 '12 at 19:13
The Riemann Zeta function $\zeta(s)=\sum_1^{\infty}n^{-s}$ generalizes your sum from reciprocal powers of $6$ to (almost) any (complex) exponent $s$ (the real part of $s$ must be greater than one for the series to converge). It is an extensively and actively studied function, quite mysterious and beloved by mathematicians. – bgins Apr 27 '12 at 19:19
@bgins The somewhat beautiful thing with Parseval here is that it gives an easy way to actually calculate $\zeta(2n)$ $n\ge1$! Other values are much harder: (1) [Apéry]:( proved $\zeta(3)$ is irrational in 1979. (2) [Zudilin]:( proved that $\zeta(5),\,\zeta(7),\,\zeta(9),\,\zeta(11)$ is irrational in 2001. – AD. Apr 27 '12 at 20:06


  1. In Parseval's formula you will square the Fourier coefficients $c_n$.

  2. Can you prove for a generic function $f$ that $$c_n(f)= j_n+\frac{t}{n}c_n(df/dx)$$ for suitable numbers $j_n$ and $t$.

2'. Can you prove for a generic function $f$ that $$c_n(f)= j_n+\frac{k_n}{n}+\frac{t}{n^2}c_n(d^2f/dx^2)$$ for suitable numbers $j_n,\,k_n$ and $t$.

2''. Can you prove for a generic function $f$ that $$c_n(f)= j_n+\frac{k_n}{n}+\frac{l_n}{n^2}+\frac{t}{n^3}c_n(d^3f/dx^3)$$ for suitable numbers $j_n,\,k_n,\,l_n$ and $t$.

I hope you see the picture.

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