How can we show that $\sum_{n=1}^{\infty}{\zeta(2n)\over a^{2n}n}=\ln\left({\pi\over a\sin({\pi\over a})}\right)$?

Question

Using Sum Calculator on $(1)$,

$$\sum_{n=1}^{\infty}{\zeta(2n)\over a^{2n}n}\tag1$$ for $a>1$

we noticed that it takes a closed form of

$$(1)=\ln\left({\pi\over a\sin({\pi\over a})}\right)\tag2$$

How can we show that whether $(1)$ is correct or not?

We can see that

$${\sin x\over x}=\prod_{n=1}^{\infty}\left(1-{x^2\over n^2\pi^2}\right)\tag3$$

It looks like that the product $(3)$ can we change into a sum, but I can't see how it is done.

Answer 1

\begin{eqnarray*} \sum_{n=1}^{\infty} \frac{ \zeta(2n)}{a^{2n} n} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{ 1}{a^{2n} n m^{2n}} = \sum_{m=1}^{\infty} \ln \left( 1 - \frac{1}{a^2 m^2} \right) \\= \ln \prod_{m=1}^{\infty} \left( 1 - \frac{1}{a^2 m^2} \right) = \ln ( \frac{a}{\pi} \sin ( \frac{ \pi}{a} ) ). \end{eqnarray*}

Answer 2

With formula $$\sum_{n=1}^\infty\zeta(2n)x^{2n}=\dfrac12\left(1-\pi x\cot\pi x\right)$$ we have $$\sum_{n=1}^\infty\zeta(2n)x^{n-1}=\dfrac12\left(\dfrac1x-\dfrac{\pi}{\sqrt{x}}\cot\pi\sqrt{x}\right)$$ after integration $$\sum_{n=1}^\infty\zeta(2n)\dfrac{x^n}{n}=\log\dfrac{\sqrt{x}}{\sin\pi \sqrt{x}}+C$$ then set $x=\dfrac{1}{a^2}$ where $C=\log\pi$ obtaining with limit as $x\to0$.