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: $$$$\cos(ax) = \dfrac{2a\sin(a\pi)}{\pi}\left( \dfrac{1}{2a^2} + \sum_{n=1}^{\infty} \dfrac{(-1)^n}{a^2-n^2} \cos(nx) \right)$$$$

b) From the preceding result show that:

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

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 $$x=\pi$$:

$$$$\cos(a\pi) = \dfrac{2a\sin(a\pi)}{\pi}\left( \dfrac{1}{2a^2} + \sum_{=1}^{\infty} \dfrac{(-1)^n}{a^2-n^2} \cos(n\pi) \right)$$$$

and after some algebra i ended up with this:

$$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\left( \dfrac{a^2}{n^2-a^2}\right)$$$$

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:

$$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \left( \dfrac{1}{1-\dfrac{a^2}{n^2}} \right)$$$$ $$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \left(\dfrac{a^2}{n^2}\right)^p$$$$ $$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \left(\dfrac{a}{n}\right)^{2p}$$$$ $$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \dfrac{1}{n^{2p}}a^{2p}$$$$ $$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \zeta(2p)a^{2p}$$$$

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.

• $a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\sum_{p=1}^{\infty} \left(\dfrac{a^2}{n^2}\right)^p$, change the order of summation then make $\zeta(2p)$ appear. Oct 21, 2020 at 7:07

We have $$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \left( \dfrac{1}{1-\dfrac{a^2}{n^2}} \right)$$$$ $$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \sum_{p=0}^{\infty} \dfrac{a^2}{n^2}\left(\dfrac{a^2}{n^2}\right)^p$$$$ $$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\sum_{p=0}^{\infty} \left(\dfrac{a^2}{n^2}\right)^{p+1}$$$$ $$$$a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\sum_{p=1}^{\infty} \left(\dfrac{a^2}{n^2}\right)^{p}$$$$ $$$$a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{n^{2p}}a^{2p}$$$$ $$$$a\pi\cot{a\pi} = 1-2 \sum_{p=1}^{\infty} \zeta(2p)a^{2p}$$$$
• I see now, i should've started the sumation at $p=0$ and after that, you can manipulate the $p$ index so it starts at 1 and then change the order of summations (as @reuns said) to make $\zeta(2p)$ appear and get the result, thanks man! Oct 21, 2020 at 11:50