Closed form of $\zeta(2n)$

Question

Is it possible to use $$\frac{\sin \pi s}{\pi s}=\prod _{n\ge 1}\left(1-\frac{s^2}{n^2}\right)$$to get a closed form for $\zeta(2n)$?

I am told that it is possible by using Taylor expansion. I can see that expanding the LHS gives $\zeta(2)$ as the coefficient of $s^2$ which makes that case easy to handle. But the other coefficients are not so straightforward. For example, is it at least possible to get the value of $\zeta(4)$ from this approach?

The $s^{2k}$ coefficient depends on each $\zeta(2n)$ with $1\le n\le k$ in a way that's fiddly to work out. It can be done, but there are better ways to get the general result. — J.G., Apr 06 '25 at 08:12

score 4 · Accepted Answer · answered Apr 06 '25 at 08:04

4

Yes it is—in fact Euler did exactly this! You can find the derivation in these solutions from an old class I taught.

answered Apr 06 '25 at 08:04

Greg Martin

92,241

4

Why did this have a downvote? – Dietrich Burde Apr 06 '25 at 10:28
Thank you very much! (+1) – Apr 06 '25 at 15:37

score 1 · Answer 2 · edited Apr 12 '25 at 02:49

If by "closed form" you mean a recursive formula or in terms of the Bernoulli numbers, the answer is yes.

Step one Take logarithm both sides

Step two Differentiate both sides

Step three write $\frac{1}{n^2-z^2}$ as the geometric series $$\frac{1}{n^2} \left(1+\frac{z^2}{n^2}+\cdots\right) $$

Step four Obtain the Taylor series $$\pi\cot \pi z=\frac{1}{z}-\sum_{n=1}^\infty 2\zeta(2n)z^{2n-1}$$

You can now use the differential relation $\cot'=-1-\cot^2$ and compare coefficients on both sides to obtain a recursive formula.

To write $\zeta$ in terms of the Bernoulli numbers it suffices to write the cotangent function in terms of $\frac{z}{e^z-1}$

Closed form of $\zeta(2n)$

2 Answers2