How to compute $\sum_{k=1}^{\infty}{(\zeta(2k)-1)}$

Question

How to compute $\sum_{k=1}^{\infty}{(\zeta(2k)-1)}$, where $\zeta(s) :=\sum_{k=1}^{\infty} \frac{1}{n^s}$ with $s>1$.

Here's my process, what am I doing wrong?:

Answer 1

Using $~\displaystyle\prod\limits_{k=1}^\infty\left(1-\left(\frac{x}{k}\right)^2\right)=\frac{\sin(\pi x)}{\pi x}~$ we have: $$\sum\limits_{k=1}^\infty\frac{x^{2k}}{2k}\zeta(2k) = -\frac{1}{2}\ln\frac{\sin(\pi x)}{\pi x}~,~~|x|<1$$

Differentiation and then substraction of $~\displaystyle\sum\limits_{k=1}^\infty x^{2k-1} = \frac{x}{1-x^2}~$ and after that the limit for $~x\uparrow 1~$ gives us the result $~\displaystyle\frac{3}{4}~$.

Hint: $~~\displaystyle\lim_{x\to 0}\left(\frac{1}{x}+\pi\cot(\pi(1-x))\right)=0$

Note: $~~$ But of course, using the result of e.g. Jam, it's easier to write

$\displaystyle\sum\limits_{n=2}^\infty\frac{1}{n^2-1} = \frac{1}{2}\sum\limits_{n=2}^\infty \left(\frac{1}{n-1} - \frac{1}{n+1}\right) = \frac{1}{2}\left(1+\frac{1}{2}\right)=\frac{3}{4}\,$ .

Answer 2

Your first error is swapping the indexes $n$ and $k$ at the start of the second line, since the $n=1$ term is $\left(\sum_{k=1}^\infty\frac{1}{1^{2k}}\right)-1=\left(\sum_{k=1}^\infty1\right)-1$, which is undefined. This is compounded by the lack of brackets, which makes it unclear what is inside the sum. Your second error is also in the second line, where you have

$$\begin{aligned} \sum_{n=1}^\infty\left(\sum_{k=1}^\infty\left(\frac1{n^2}\right)^k-1\right)&=\sum_{n=1}^\infty\left(\sum_{\color{red}{k=2}}^\infty\left(\frac1{n^2}\right)^k-1+\left(\frac1{\color{red}{1}^2}\right)^k\right) \end{aligned} $$

Ignoring that the sum is undefined due to the $n=1$ term, this is also incorrect since you are accounting for a term in $n$ by changing the index in $k$. However, it is clear that you are trying to take out a term of the series, which is a good tactic but should be done as $\sum_{i=1}^Na_i=a_1+a_2+\ldots+a_N=\color{red}{a_1}+\sum_{\color{red}{i=2}}^Na_i$. The index you change should reflect the index you substitute.

You can mitigate these two errors by first dealing with the $n=1$ term before swapping the indexes.

$$\require{cancel}\begin{aligned} \sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac1{n^{2k}}-1\right) &= \sum_{k=1}^\infty\left(\cancel{\frac{1}{1^{2k}}}+\frac{1}{2^{2k}}+\ldots\cancel{-1}\right) \\ &=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac{1}{n^{2k}} \end{aligned} $$

At which point, you could swap the indexes, use your geometric series argument from the third line and complete the proof.

$$\begin{aligned} &=\sum_{n=2}^{\infty}\frac{1}{n^2-1} \end{aligned} $$

You should try not to assume that a rearrangement is convergent without justification since it is not a given that $\sum_n$ and $\sum_k$ could be swapped or that the geometric series would even converge. You could also try to improve your notation since your $1$s look like $7$s.