Question
$$\zeta(k)=1+\dfrac{1}{2^k}+\dfrac{1}{3^k}+\cdots+\dfrac{1}{n^k}+\cdots$$ Prove that : $$\sum_ {k=1}^{\infty}\dfrac{\zeta(2k)-\zeta(3k)}{k}=\ln\left(\frac{2\cosh\left(\sqrt{3}\pi/2\right)}{3\pi}\right).$$
We have
$$\begin{align}
\sum\limits_{k = 1}^\infty {\frac{{\zeta \left( {2k} \right) – \zeta \left( {3k} \right)}}{k}} &= \sum\limits_{k = 1}^\infty {\frac{1}{k}\left( {\sum\limits_{n = 2}^\infty {\left( {\frac{1}{{{n^{2k}}}} – \frac{1}{{{n^{3k}}}}} \right)} } \right)}\\&= \sum\limits_{n = 2}^\infty {\sum\limits_{k = 1}^\infty {\left( {\frac{1}{{k{n^{2k}}}} – \frac{1}{{k{n^{3k}}}}} \right)} }\\&= \sum\limits_{n = 2}^\infty {\left( { – \ln \left( {1 – \frac{1}{{{n^2}}}} \right) + \ln \left( {1 – \frac{1}{{{n^3}}}} \right)} \right)}\\&
= – \ln \frac{3}{2} + \sum\limits_{n = 1}^\infty {\ln \left( {\frac{{{n^2} + n + 1}}{{{n^2} + n}}} \right)}\\&= – \ln \frac{3}{2} + \ln \left( {\prod\limits_{n = 1}^\infty {\frac{{\left( {n – {z_1}} \right)\left( {n – {z_2}} \right)}}{{n\left( {n + 1} \right)}}} } \right)
\end{align}$$
Best Answer
Let $j:=\frac{-1+i\sqrt3}2.$ You proved that $$\sum_{k = 1}^\infty\frac{\zeta(2k)- \zeta(3k)}k=\ln\prod_{n =2}^\infty\frac{(n-j)(n+1+j)}{n(n + 1)}.$$ Now, $$\begin{align}\prod_{n =2}^\infty\frac{(n-j)(n+1+j)}{n(n + 1)}&=\frac1{(1-j)(1+j)(1+j/2)}\prod_{n =1}^\infty\left(1-\frac{j^2}{n^2}\right)\\ &=\frac{-2j}3\frac{\sin(\pi j)}{\pi j}\\ &=\frac{2\cosh\left(\pi\sqrt3/2\right)}{3\pi}. \end{align}$$ See also https://oeis.org/A073017.