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Can someone explain how one can go about in finiding certain values of the function?

$\zeta(2)=?$

I have tried using fourier analysis; however it seems overwhelmingly difficult.

BobKim
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1 Answers1

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$\zeta(s)$ is in generally pretty hard to compute. There is a formula for $\zeta(2n)$, for natural $n$, involving Bernoulli numbers, it is given by: $$\zeta(2n) = \frac{(-1)^{n+1}B_{2n}(2\pi)^{2n}}{2(2n)!} $$

Where $B_k$ is the $k$-th Bernoulli number. There is also a very nice result by Apéry's that $\zeta(3)$ is irrational.

Other than those, closed forms are very elusive for $\zeta(s)$.

doppz
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