<s>[INST] How to compute $\sum_{k=1}^{\infty}{(\zeta(2k)-1)}$ 
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?:

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