I found an interesting identity regarding the Riemann zeta function:
$$ \sum_{k=1}^{\infty}(\zeta(2k)-1)x^{2k}= \frac{x^2}{1-x^2} + \frac{1}{2} (1-\pi x \cot(\pi x) )$$
but to arrive at this identity, it is necessary to use:
$$\sum_{k=1}^{\infty}x^{2k} = \frac{x^2}{1-x^2} $$
where $|x^2|<1$, because it is a geometric series.
On the other hand, pluging in $x=1$ gives us:
$$ \sum_{k=1}^{\infty}(\zeta(2k)-1)=\frac{3}{4}$$
which is another identity. Why can we do that? Can we also use $x=i$?
Because of the singularity, do not simplify too much and stay with $$f(x)=\sum_{k=1}^{\infty}(\zeta(2k)-1)x^{2k}=\frac{1-3 x^2-\pi x \left(1-x^2\right) \cot (\pi x)}{2 \left(1-x^2\right)}$$ $$f(i)=1-\frac{\pi}{2} \coth (\pi )$$