Show that $a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \zeta(2p)a^{2p}$

Question

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.

Answer 1

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}