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Can anybody pass me on a good source to see the steps in proving, \begin{equation} \zeta(2n) = \frac{(-1)^{k-1}B_{2k}(2\pi)^{2k}}{2(2k)!} \end{equation}

I know how we start by looking at the product of sine and use the generatinf function for the Bernoulli numbers to connect them. I am finding it hard to find a source that doesn't just assume the result or say that it is fairly trivial.

Any help would be appreciated, Thanks

metamorphy
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Sean
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    It's often shown when deriving the partial fraction decomposition of the cotangent. From the partial fraction decomposition, you obtain the Laurent series with the $\zeta(2n)$ coefficients by expanding the fractions in geometric series and changing the order of summation, and on the other hand, you get the Laurent series with the Bernoulli numbers almost by definition of the Bernoulli numbers. – Daniel Fischer Aug 02 '15 at 20:57
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    By my memory, I'd say that it has already been asked, but I am not able to find a duplicate at the moment. – Jack D'Aurizio Aug 02 '15 at 21:16
  • Do you know residue calculus? If you do, consider the meromorphic function $f(s)=\frac{1}{s^n(e^s-1)}$ for integer $n>=2$ Show that a circular contour integral tends to zero as the radius tends to infinity. Hence, the sum of the residues of the function is zero. At all non-zero integers the residues will add up to the zeta function multiplied by all the rest in your identity, at zero the residue will be in terms of a Bernoulli number... – Asier Calbet Aug 03 '15 at 08:34
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    @JackD'Aurizio http://math.stackexchange.com/questions/1322604/ways-to-prove-eulers-formula-for-zeta2n/1322623#1322623 :) – Noam Shalev - nospoon Aug 03 '15 at 08:40

2 Answers2

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There is a nice proof in Remmert's book "Funktionentheorie" using residue calculus. Also, Tom M. Apostol has given a short elementary proof in The American Mathematical Monthly 1973 (JSTOR, Google), based on quadratic cotangent identities. The article also surveys other proofs of this famous identity, and gives many useful references.

Martin Sleziak
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Dietrich Burde
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This happens to have been the topic of an article in the May 2015 American Mathematical Monthly. You can get access here.

Edit: instead of me sumerizing the article, you can find it here for free (thanks to Raymond Manzoni for the link).

user2520938
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