Show that $a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \zeta(2p)a^{2p}$ I'm trying to solve the problem 14.3.9 (Applications of Fourier Series) from Arfken's Mathematical Methods For Physicists:
a) Show that the fourier expansion of $\cos(ax)$ is:
\begin{equation}
\cos(ax) = \dfrac{2a\sin(a\pi)}{\pi}\left( \dfrac{1}{2a^2} + \sum_{n=1}^{\infty} \dfrac{(-1)^n}{a^2-n^2} \cos(nx) \right)
\end{equation}
b) From the preceding result show that:
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \zeta(2p)a^{2p}
\end{equation}
where $ \zeta(2p)$ is the riemann zeta function $ \zeta(2p) = \sum_{n=1}^{\infty} \dfrac{1}{n^{2p}}$
I´ve already solved part a), but im stuck on part b), what i did was the following, first i evalueted $\cos(ax)$ at $x=\pi$:
\begin{equation}
\cos(a\pi) =  \dfrac{2a\sin(a\pi)}{\pi}\left( \dfrac{1}{2a^2} + \sum_{=1}^{\infty} \dfrac{(-1)^n}{a^2-n^2} \cos(n\pi) \right)
\end{equation}
and after some algebra i ended up with this:
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\left( \dfrac{a^2}{n^2-a^2}\right)
\end{equation}
which is the part i'm stuck, i'm not sure how to relate this last expression with $\sum_{p=1}^{\infty} \zeta(2p)a^{2p}$, i was thinking to use the geometric series and tried something like this:
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \left( \dfrac{1}{1-\dfrac{a^2}{n^2}} \right)
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \left(\dfrac{a^2}{n^2}\right)^p
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \left(\dfrac{a}{n}\right)^{2p}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \dfrac{1}{n^{2p}}a^{2p}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \zeta(2p)a^{2p}
\end{equation}
but i get a different result and i don't know in which part i was wrong or if i'm missing something. Any help would be appreciated, thanks.

We have \begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \left( \dfrac{1}{1-\dfrac{a^2}{n^2}} \right)
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}  \sum_{p=0}^{\infty} \dfrac{a^2}{n^2}\left(\dfrac{a^2}{n^2}\right)^p
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\sum_{p=0}^{\infty} \left(\dfrac{a^2}{n^2}\right)^{p+1}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\sum_{p=1}^{\infty} \left(\dfrac{a^2}{n^2}\right)^{p}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{n^{2p}}a^{2p}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2 \sum_{p=1}^{\infty} \zeta(2p)a^{2p}
\end{equation}
