How can I prove that $\zeta(2n) = (-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$?

Asked Dec 20 '15 at 19:49

Active Dec 20 '15 at 20:51

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I tried to prove this with Valli's equation: $\frac{sinx}{x} = \prod(1-\frac{x^{2}}{(n\pi)^{2}})$ and use $\frac{d(ln(sin(x)))}{dx} = i + \frac{2i}{e^{2ix}-1}$.

Maybe it's better to use Taylor's transform?

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

edited Dec 20 '15 at 20:51

Silvia Ghinassi

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asked Dec 20 '15 at 19:49

openspace

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See also here. – Dietrich Burde Dec 20 '15 at 19:51
@DietrichBurde thanks a lot! – openspace Dec 20 '15 at 19:52
1

You are welcome - the last link has an article with many nice proofs. – Dietrich Burde Dec 20 '15 at 19:52

How can I prove that $\zeta(2n) = (-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$?

0 Answers0