Summation with Zeta function

Question

I'm currently studiyng Zeta function and I don't understand this identity:

$$\sum_{n=1}^\infty x \sum_{p=0}^\infty \frac{x^{2p}}{(n\pi)^{2p+2}} = \sum_{p=1}^\infty \pi^{-2p}\zeta(2p)x^{2p-1} $$

I think the authors are switching the sum symbols but everything I try fails. Thanks for your help.

Answer 1

Yes, changing summation is involved (along with reindexing $p$ to start with $p = 1$):

\begin{align} \sum_{n = 1}^\infty x \sum_{p = 0}^\infty \frac{x^{2p}}{(n\pi)^{2p+2}} &= \sum_{n = 1}^\infty \sum_{p = 0}^\infty \frac{x^{2p+1}}{(n\pi)^{2p+2}}\\ &= \sum_{n = 1}^\infty \sum_{p = 1}^\infty \frac{x^{2p-1}}{(n\pi)^{2p}}\\ &= \sum_{p = 1}^\infty \pi^{-2p} x^{2p-1} \sum_{n = 1}^\infty n^{-2p}\\ &= \sum_{p = 1}^\infty \pi^{-2p}\zeta(2p) x^{2p-1}. \end{align}