Zeta for Even integers proof (Bernoulli numbers)

$$\zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$

$$\textit{proof}$$

We start by the product formula of the sine function

$$\frac{\sin(z)}{z} = \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$

Take the logarithm to both sides

$$\log(\sin(z)) – \log(z) = \sum_{n=1}^\infty \log \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$

By differentiation with respect to $z$

$$\cot(z) -\frac{1}{z} = -2\sum_{n=1}^\infty \frac{ \frac{z}{n^2 \, \pi^2 }}{1-\frac{z^2}{n^2 \, \pi^2}}$$

By simple algebraical manipulation we have

$$z\cot(z) = 1-2\sum_{n=1}^\infty \frac{z^2}{ n^2 \, \pi^2} \left(\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}} \right)$$

Now using the power series expansion

$$\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}}= \sum_{k=0}^\infty \frac{1}{n^{2k}\, \pi^{2k}} z^{2k} \,\,\,,\,|z|< \pi \, n $$

$$\frac{z^2}{ n^2 \, \pi^2}\left(\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}} \right) =\sum_{k=0}^\infty\frac{1}{n^{2k+2}\, \pi^{2k+2}} z^{2k+2}= \sum_{k=1}^\infty\frac{1}{n^{2k}\, \pi^{2k}} z^{2k}$$

So the sums becomes

$$z\cot(z) = 1-2\sum_{n=1}^\infty \sum_{k = 1}^\infty \frac{1}{n^{2k}} \, \frac{z^{2k}}{\pi^{2k}}$$

Now if we invert the order of summation we have

$$z\cot(z) = 1-2\sum_{k=1}^\infty \sum_{n = 1}^\infty \frac{1}{n^{2k}} \, \frac{z^{2k}}{\pi^{2k}} = 1-2\sum_{k=1}^\infty \frac{\zeta(2k) }{\pi^{2k}}z^{2k} $$

Euler didn’t stop here, he used power series for \(z\cot(z) \) using the Bernoulli numbers.

Start by the equation

$$\frac{x}{e^x-1} =\sum_{k=0}^\infty \frac{B_k}{k!}x^k$$

By putting $x=2iz $ we have

$$\frac{2iz}{e^{2iz}-1} = \sum_{k= 0}^\infty \frac{B_k}{k!}{(2iz)}^k$$

Note that \( B_{2k+1}=0 \)

$$z \cot(z) = 1- \sum_{k=1}^\infty(-1)^{k-1} B_{2k} \frac{2^{2k}}{(2k)!}z^{2k} $$

The result is immediate by comparing the two different representations.

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