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I have been solving problems using a "Potentially Helpful Formulas" sheet from my esteemed math professor. i want to solve for:

$\sum_{n=1}^{\infty} \dfrac{1}{n^4} =$ ?

On my formula sheet i have:

$\sum_{n=1}^{\infty} \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$

Which looks similar to what i want, but not quite. Might someone guide me towards the solution (or provide the solution with explanation?


marked as duplicate by Mark Bennet, user147263, Mathmo123, Rick Decker, apnorton Aug 6 '14 at 0:12

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.


This is just a sketch of one of many possible proofs.

Step1. Prove that over the interval $[0,2\pi]$, the function: $$f(x)=\sum_{n=1}^{+\infty}\frac{\cos(nx)}{n^2}$$ is a second degree-polynomial whose graphics passes through the points: $$(0,\pi^2/6),\quad (\pi,-\pi^2/12),\quad (2\pi,\pi^2/6).$$

Step2. Deduce from Lagrange interpolation that: $$ f(x) = \frac{\pi^2}{6}-\frac{x(2\pi-x)}{4}.$$

Step3. Apply Parseval's identity to $f(x)$: $$\int_{0}^{2\pi}f(x)^2\,dx = \pi\sum_{n=1}^{+\infty}\frac{1}{n^4}.$$

Step4. Prove, through the second step, that: $$\int_{0}^{2\pi}f(x)^2\, dx = \frac{\pi^5}{90}.$$


$$\zeta(4)=\sum_{n=1}^{+\infty}\frac{1}{n^4} = \frac{\pi^4}{90}.$$


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