Summation with Riemann Zeta Function

Question

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 $$\sum \limits_{k=1}^{\infty }\frac{\zeta(2k)-1}{k}$$

Answer 1

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 $c=\frac{1}{j^2}$ $$\displaystyle \sum_{i=2}^{\infty} \frac{1}{j^{2i}\times i}=-\ln \left(1-\frac{1}{j^2}\right)$$ $$\displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k} = -\sum_{n=2}^{\infty} \ln \left(1-\frac{1}{n^2}\right)=-\ln\left(\prod_{n=2}^{\infty} \left(1-\frac{1}{n^2}\right)\right)$$ $$\displaystyle \prod_{n=2}^{\infty} \left(1-\frac{1}{n^2}\right)=\prod_{n=2}^{\infty}\left(1-\frac{1}{n}\right) \times \prod_{n=2}^{\infty}\left(1+\frac{1}{n}\right)=\frac{1}{2}$$ $$\therefore\displaystyle \sum_{k=1}^{\infty} \frac{\zeta(2k)-1}{k}=\ln(2)$$

Answer 2

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