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.


  • $\begingroup$ 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!

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  • $\begingroup$ 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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