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.
Thanks
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!
