A zeta series

Evaluate the series:

$$\sum_{n=1}^{\infty} \frac{\zeta(2n)-\zeta(3n)}{n}$$

Solution

We are needing the following product as a lemma:

Lemma: It holds that:

$$\prod_{k=2}^{\infty} \left ( 1- \frac{z^n}{k^n} \right )= \prod_{k=0}^{n-1} \frac{1}{\Gamma \left ( 1-z\exp \left [ \left ( 2\pi k i /n \right ) \right ] \right )}$$

Proof: A proof may be found here

Thus for the initial question:

\begin{align*}
\sum_{n=1}^{\infty}\frac{\zeta(2n)-\zeta(3n)}{n} &=\sum_{n=1}^{\infty} \frac{1}{n}\left [ \sum_{k=1}^{\infty} \frac{1}{k^{2n}} - \sum_{k=1}^{\infty} \frac{1}{k^{3n}} \right ] \\
 &=\sum_{n=1}^{\infty} \frac{1}{n}\sum_{k=2}^{\infty} \left [ \frac{1}{k^{2n}} - \frac{1}{k^{3n}} \right ] \\
 &= \sum_{k=2}^{\infty} \left [ \sum_{n=1}^{\infty} \frac{1}{n}\left ( \frac{1}{k^2} \right )^n - \sum_{n=1}^{\infty} \frac{1}{n} \left ( \frac{1}{k^3} \right )^n \right ]\\
 &= \sum_{k=2}^{\infty} \left [ - \log \left ( 1- \frac{1}{k^2} \right ) + \log \left ( 1 - \frac{1}{k^3} \right ) \right ]\\ 
 &= - \log \prod_{k=2}^{\infty} \left ( 1- \frac{1}{k^2} \right ) + \log \prod_{k=2}^{\infty} \left ( 1-\frac{1}{k^3} \right ) \\
 &=\log 2 + \log \left ( \frac{\cosh \left ( \frac{\sqrt{3}\pi}{2} \right )}{3\pi} \right ) \\
 &= \log \left( \frac{2 \cosh \left(\frac{\sqrt{3} \pi}{2}\right)}{3 \pi} \right)
\end{align*}