TITLE: Summation with Riemann Zeta Function
QUESTION [5 upvotes]: 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 $$\sum \limits_{k=1}^{\infty }\frac{\zeta(2k)-1}{k}$$

REPLY [2 votes]: 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
$$ \sum_{k=1}^{\infty} \frac{1}{k} \left( \zeta(2k)-1 \right) = \log{2}. $$