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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?

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Good luck with this. I am curious myself. – 000 Nov 20 '12 at 1:02
I have proposed some ideas concerning your calculation at this link. – Marko Riedel Dec 1 '12 at 9:33
@MarkoRiedel I know some proofs that evaluate $\zeta(2k)$ in terms of $B_{2k}$. For instance, page $15$ here. In this question, I am interested in whether this identity for Bernoulli numbers is already known or not. – user17762 Dec 1 '12 at 19:08

As far as I can tell, you have already proven your claim; you started with a function with the appropriate value for the coefficient of interest and then found an expression for that coefficient. Consequently that expression must be a Bernoulli number by construction. If for some reason you are after an alternative proof that the coefficient of $z^{2n}$ in the function:



$f \left( z \right) =\prod _{k=0}^{n-1}\mathrm{sinc} \left( z \exp \left( -\dfrac{k \pi i}{n} \right)\right) $,


is equal to the Bernoulli number $B(2n)$ then here's one...

First take the logarithm of $f(z)$:

$\ln(f \left( z \right)) =\sum _{k=0}^{n-1}\ln\left(\mathrm{sinc} \left( z \exp \left(- \dfrac{k \pi i}{n} \right)\right)\right)$.

Then use the product expansion of the sinc function:

$\mathrm{sinc} \left( z \exp \left( -\dfrac{k \pi i}{n} \right)\right)=\prod _{q=1}^{\infty}\left(1-\left(\dfrac{z}{{\pi}q}\right)^2\exp \left(- \dfrac{2k \pi i}{n} \right)\right)$

to obtain:

$\ln(f \left( z \right)) =\sum _{k=0}^{n-1}\sum _{q=1}^{\infty}\ln\left(1-\left(\dfrac{z}{{\pi}q}\right)^2\exp \left(- \dfrac{2k \pi i}{n} \right)\right)$.

then the series expansion of the logarithm:

$\ln\left(1-\left(\dfrac{z}{{\pi}q}\right)^2\exp \left(- \dfrac{2k \pi i}{n} \right)\right)=-\sum _{r=1}^{\infty}\left(\dfrac{z}{{\pi}q}\right)^{2r}r^{-1} \exp \left(- \dfrac{2kr \pi i}{n} \right)$

to get:

$\ln(f \left( z \right)) =-\sum _{k=0}^{n-1}\sum _{q=1}^{\infty}\sum _{r=1}^{\infty}\left(\dfrac{z}{{\pi}q}\right)^{2r}r^{-1} \exp \left(- \dfrac{2kr \pi i}{n} \right)$.

$\ln(f \left( z \right)) =-\sum _{q=1}^{\infty}\sum _{r=1}^{\infty}\left(\dfrac{z}{{\pi}q}\right)^{2r}r^{-1} \sum _{k=0}^{n-1}\exp \left(- \dfrac{2kr \pi i}{n} \right)$.

Note the Kronecker delta popping up that will only pick out $r$ when it's a multiple (denoted $l$) of $n$:

$\sum _{k=0}^{n-1}\exp \left(- \dfrac{2kr \pi i}{n} \right)=n\delta_{r,nl}$,

so we may multiply by $n$, replace $r$ with $nl$ and sum over $l$:

$\ln(f \left( z \right)) =-\sum _{q=1}^{\infty}\sum _{l=1}^{\infty}\left(\dfrac{z}{{\pi}q}\right)^{2nl}l^{-1} $.

Now recognise the sum over $q$:

$\sum _{q=1}^{\infty}\dfrac{1}{q^{2nl}}=\zeta(2nl)$

and so it follows:

$\ln(f \left( z \right)) =-\sum _{l=1}^{\infty}\zeta(2nl)\left(\dfrac{z}{{\pi}}\right)^{2nl}l^{-1} $.

Now take the exponential:

$f \left( z \right) =\exp\left(-\sum _{l=1}^{\infty}\zeta(2nl)z^{2nl}{\pi}^{-2nl}l^{-1} \right)$,

$f \left( z \right) =\prod _{l=1}^{\infty}\exp\left(-\zeta(2nl)z^{2nl}{\pi}^{-2nl}l^{-1} \right)$,

$f \left( z \right) =\prod _{l=1}^{\infty}\sum _{u=0}^{\infty}\left(-\zeta(2nl)z^{2nl}{\pi}^{-2nl}l^{-1} \right)^u(u!)^{-1}$.

We are interested in the coefficient of $z^{2n}$ and this will not receive any contribution from $1<l$ or $1<u$ and thus:

$f \left( z \right) =1-\zeta(2n)z^{2n}{\pi}^{-2n}+O(z^{4n})$,

and consequently:

$g(z)=2(-1)^n(2n)!2^{-2n}f(z)= 2(-1)^n(2n)!2^{-2n}-2(-1)^n(2n)!2^{-2n}\zeta(2n)z^{2n}{\pi}^{-2n}+O(z^{4n})$.

From which it follows that the coefficient of $z^{2n}$ is:


As you have supplied an alternative expression for the same coefficient your expression must also equal $B(2n)$.

It seems like an interesting expression, do you have an application for it? Can you use it to calculate Fourier coefficients in products of series for example or something like that?

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Note the $(-1)^{n}$ in $g(z)$ is absent from your starting function; I introduced this as I think you may have missed it somewhere along the line, but correct me if I am wrong. – Graham Hesketh Apr 28 '13 at 11:27
Thanks. Let me look into this and get back to you. – user17762 Apr 28 '13 at 19:00
You can get displayed equations (centred and less squashed) by using double dollar signs instead of single dollar signs. – joriki May 29 '13 at 10:58
Cheers, I realised that now. – Graham Hesketh May 29 '13 at 11:06

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