Summation with Riemann Zeta Function So the Riemann zeta function $\zeta(s)$ is commonly defined as $\sum \limits_{n=1}^{\infty} n^{-s}$
Now, suppose that $a_k=\zeta (2k).$ How can I find the value of $$\sum \limits_{k=1}^{\infty }\frac{\zeta(2k)-1}{k}$$
 A: Note that:
$$\displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k}=\sum_{i=2}^{\infty} \sum_{j=1}^{\infty} \frac{1}{j \times i^{2j}}$$
Also note that:
$$\displaystyle \int_{0}^{c} \frac{1}{1-x} dx=-\ln(1-c)=\sum_{i=1}^{\infty} \frac{c}{i}$$
Now, if we substitute $c=\frac{1}{j^2}$
$$\displaystyle \sum_{i=2}^{\infty} \frac{1}{j^{2i}\times i}=-\ln \left(1-\frac{1}{j^2}\right)$$
$$\displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k} = -\sum_{n=2}^{\infty} \ln \left(1-\frac{1}{n^2}\right)=-\ln\left(\prod_{n=2}^{\infty} \left(1-\frac{1}{n^2}\right)\right)$$
$$\displaystyle \prod_{n=2}^{\infty} \left(1-\frac{1}{n^2}\right)=\prod_{n=2}^{\infty}\left(1-\frac{1}{n}\right) \times \prod_{n=2}^{\infty}\left(1+\frac{1}{n}\right)=\frac{1}{2}$$
$$\therefore\displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k}=\ln(2)$$
A: These sorts of sums are all done in the same way: you expand the zeta function using the Dirichlet series, then interchange the order of summation. Viz:
$$\begin{align}
\sum_{k=1}^{\infty} \frac{1}{k} \left( \zeta(2k)-1 \right) &= \sum_{k=1}^{\infty} \frac{1}{k} \left( \sum_{n=2}^{\infty} \frac{1}{n^{2k}} \right) \\
&= \sum_{n=2}^{\infty} \sum_{k=1}^{\infty} \frac{(n^{-2})^k}{k} \\
&= \sum_{n=2}^{\infty} -\log{\left(1-\frac{1}{n^2}\right)}
\end{align}$$
Now, the annoying bit is to set this up to telescope: write
$$ -\log{\left(1-\frac{1}{n^2}\right)} = -\log{(n+1)}+ 2\log{n} - \log{(n-1)} $$
It's fairly clear what's going to happen now: you get a series of terms of the form
$$ \dotsb + (-\log{n}+ 2\log{(n+1)} - \log{(n-1)}) \\
 + (-\log{(n+1)}+ 2\log{n} - \log{(n-1)}) \\
 + (-\log{(n+2)}+ 2\log{(n+1)} - \log{n}) + \dotsb, $$
and clearly everything cancels but the first couple (oh, by the way, convergence? The bracketed terms are bounded by $2/n^2$, if you're careful), and you find
$$ \sum_{k=1}^{\infty} \frac{1}{k} \left( \zeta(2k)-1 \right) = \log{2}. $$
