The Basel Problem II

Just like $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}=\zeta(2)=\dfrac{\pi^2}{6}$ is a well-known result, we also have,

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

A general result that can be derived is,

$\sum_{n=1}^\infty \frac{1}{n^{2k}}=\zeta(2k)=\frac{(-1)^{k-1}(2^{2k}B_{2k}\pi^{2k})}{2\times(2k)!}$

where $B_{2k}$ are Bernoulli numbers (some may use the notation $B_k$ since these numbers are defined only for even indexes. It all means the same).

Note: The general result is really difficult to prove (atleast for me).

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Now, coming onto the proof part, there are a lot of proofs for this, some using Fourier Trigonometric Series, some using contour integrals and even some mimicking Euler's method of finding $\zeta(2)$ to find $\zeta(4)$ by considering roots different from what Euler considered.

Here 's the link to the entire discussion. I think giving the link to the discussion is better than posting just a single proof here.