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Sometime back I made a claim here that the proof for $\zeta(4)$ can be extended to all even numbers.

I tried doing this but I face a stumbling block.

Let me explain the problem in detail here. I was trying to mimic Euler's proof for the Basel problem to evaluate $\zeta$ at all even integers and prove that $$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n} $$

Euler looks at the function whose zeros are at $\pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots$.

In general, to evaluate $\zeta(2n)$, let us look at a function whose roots are $$\pm \xi_0 \pi, \pm \xi_1 \pi, \ldots \pm \xi_{n-1} \pi; \pm \xi_0 2\pi, \pm \xi_1 2\pi, \ldots, \pm \xi_{n-1} 2\pi; \pm \xi_0 3\pi, \pm \xi_1 3\pi, \ldots, \pm \xi_{n-1} 3\pi; \ldots$$ where $\xi_k = \exp \left( \dfrac{k \pi i}{n} \right)$, $k \in \{0,1,2,\ldots,n-1\}$.

Let $$p_{2n}(z) = \left(1 - \left(\frac{z}{\pi}\right)^{2n} \right) \times \left(1 - \left(\frac{z}{2 \pi}\right)^{2n} \right) \times \left(1 - \left(\frac{z}{3 \pi}\right)^{2n} \right) \times \cdots$$

The coefficient of $z^{2n}$ in $p_{2n}(z)$ is $$- \dfrac1{\pi^{2n}}\sum_{k=1}^{\infty} \dfrac1{k^{2n}} = - \dfrac{\zeta(2n)}{\pi^{2n}}$$

It is not hard to guess that $p_{2n}(z)$ is same as $$\prod_{k=0}^{n-1} \dfrac{\sin(z/\xi_k)}{(z/\xi_k)} = \exp \left( \dfrac{(n-1) \pi i}2\right) \left(\dfrac{\sin(z/\xi_0) \sin(z/\xi_1) \sin(z/\xi_2) \ldots \sin(z/\xi_{n-1})}{z^{n}} \right)$$

Now from the power series expansion of $\sin(z)$, we get that $$\dfrac{\sin(z/\xi_k)}{z/\xi_k} = \left(\sum_{\ell=0}^{\infty} \dfrac{(-1)^{\ell} z^{2 \ell}}{(2 \ell+1)! \xi_k^{2\ell}} \right)$$

Hence, we get that $$\prod_{k=0}^{n-1} \dfrac{\sin(z/\xi_k)}{(z/\xi_k)} = \prod_{k=0}^{n-1} \left(\sum_{\ell=0}^{\infty} \dfrac{(-1)^{\ell} z^{2 \ell}}{(2 \ell+1)! \xi_k^{2\ell}} \right)$$

The coefficient of $z^{2n}$ in the product is given by $$c(n) = \sum_{\overset{\ell_0 + \ell_1 + \cdots + \ell_{n-1} = n}{\ell_k \geq 0}} \prod_{k=0}^{n-1} \left( \dfrac{(-1)^{\ell_k}}{(2 \ell_k+1)! \xi_k^{2 \ell_k}} \right)$$

Assuming whatever I have done so far is correct, to complete the proof and conclude that $$\zeta(2n) = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n} $$ I need to prove the following claim

Claim: $$\color{red}{B_{2n} = \dfrac{(2n)!}{2^{2n-1}}\sum_{\overset{\ell_0 + \ell_1 + \cdots + \ell_{n-1} = n}{\ell_k \geq 0}} \prod_{k=0}^{n-1} \left( \dfrac{1}{(2 \ell_k+1)! \xi_k^{2 \ell_k}} \right)}$$ where $\xi_k = \exp \left( \dfrac{k \pi i}{n} \right)$.

Is the above expression of the Bernoulli numbers a known result? If so, could someone point me to a proof of this result?

  • 0
    Good luck with this. I am curious myself.2012-11-20
  • 0
    I have proposed some ideas concerning your calculation at this [link](http://math.stackexchange.com/questions/248413/zeta-function-values-in-terms-of-bernoulli-numbers).2012-12-01
  • 0
    @MarkoRiedel I know some proofs that evaluate $\zeta(2k)$ in terms of $B_{2k}$. For instance, page $15$ [here](http://www.mat.uab.cat/matmat/PDFv2009/v2009n06.pdf). In this question, I am interested in whether this identity for Bernoulli numbers is already known or not.2012-12-01

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