Question on an identity the representation of the Riemann Zeta function using Bernoulli numbers

Question

I'm reading Serre's A Course in Arithmetic. In Section 4.1, we are showing that the Riemann Zeta function has a representation using the Bernoulli numbers.

The trick was to take the logarithmic derivative of $$ \sin(z) = z \prod\limits_{n = 1}^{\infty} \left(1 - \frac{z^2}{n^2 \pi^2} \right). $$

I did this and got $z \cot z = 1 + 2 \sum\limits_{n = 1}^{\infty} \frac{z^2}{z^2 - n^2 \pi^2}$, as written in Serre.

But now Serre claims that $$ z \cot z = 1 + 2 \sum_{n = 1}^{\infty} \frac{z^2}{z^2 - n^2 \pi^2} = 1 - 2 \sum_{n = 1}^{\infty} \sum_{k = 1}^{\infty} \frac{z^{2k}}{n^{2k} \pi^{2k}}. $$

How did we get the last equality?

I thought there was some geometric series tricks going on but I failed to discover the trick after a while of hard trying. Thanks for any help!

Answer 1

Just a geometric series: $$\frac{z^2}{z^2-n^2\pi^2} =-\frac{z^2}{n^2\pi^2}\frac1{1-z^2/(n^2\pi^2)} =-\frac{z^2}{n^2\pi^2}\sum_{j=0}^\infty\frac{z^{2j}}{n^{2j}\pi^{2j}} =-\sum_{k=1}^\infty\frac{z^{2k}}{n^{2k}\pi^{2k}} $$ at least if $|z|<\pi$.

Answer 2

Note that and using some of the below, you should be ale to recover the steps in the last line above. \begin{align} 1+2z^2\sum_{n=1}^{\infty} \frac{1}{z^2-n^2}&= 1-2z^2 \sum_{n=1}^{\infty}\frac{1}{n^2}\frac{1}{1-z^2/n^2} \\ &=1-2z^2\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=0}^{\infty}\left(\frac{z}{n}\right)^{2k} \\ &= 1-2 \sum_{n=1}^{\infty}z^{2k+2}\sum_{n=1}^{\infty}\frac{1}{n^{2k+2}} \\ &=1-2 \sum_{\text{even}\, k \geq 2}\zeta(k)z^k \end{align}