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How we can do this sum? $$ \sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{90} $$

I know that we could possibly use a Fourier series decomposition however I don't know what function to start with. I know it is also equal to $\zeta(4)$ but a proof involving this is tough for me since they don't teach us these at school. Thanks.



marked as duplicate by 6005, user61527, user91500, Belgi, Pedro Tamaroff calculus May 14 '14 at 4:13

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  • $\begingroup$ Try finding the Fourier series of $f(x) = (\pi - |x|)^2$ on $[-\pi, \pi)$. $\endgroup$ – Paul Hurst May 14 '14 at 4:06

By this formula derived by Euler:

$$\zeta(2n) = (-1)^{n+1} \frac{B_{2n}(2\pi )^{2n}}{2(2n)!}$$

when you let $n=2$ you have:

$$\zeta(2\cdot 2) = \sum_{n=1}^\infty \frac{1}{n^4} = 1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + ... = (-1)^{2+1} \frac{B_{2\cdot 2}(2\pi )^{2\cdot 2}}{2(2\cdot 2)!} = \frac{\pi^4}{90}$$

How to derive this formula? Long story, but you start with an infinite product for the sine function:

$$\sin(x) = x\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$

*this sine infinite product is intuitive but to prove it you can use fourier series.

Take the logarithm in both sides, derive, and then find an infinite series for the $\cot$ function.

$$x \cot x = 1 - 2\sum_{n=1}^{\infty} \left(\zeta(2n)\frac{x^{2n}}{\pi^{2n}}\right)$$

Use another $\cot$ series involving Bernoulli numbers:

$$x \cot x = 1 - 2\sum_{n=1}^\infty \left(\frac{B_{2n}}{(2n)!}\left(-\frac{1}{2}\right)(2ix)^{2n}\right)$$

And then by euqating the two series and extracting its coefficients, you'll have that Euler's formula. This guy has an awesome playlist with everything you have to know. Enjoy!

  • $\begingroup$ nobody liked my answer i'm gonna cry :'( $\endgroup$ – Lucas Zanella May 14 '14 at 4:40
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    $\begingroup$ I upvoted you to acknowledge your hard work $\endgroup$ – DeepSea May 14 '14 at 4:57

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