From the definition of $\zeta(z):= \sum_{k=1}^\infty \tfrac{1}{k^z}$ for $\mathrm{Re}(z)>1$ it is obvious that $\zeta(2k)\downarrow 1$ as $k \rightarrow \infty$. I am interested in the "true" speed of this convergence. I know that e.g. $\sum_{k=1}^\infty (\zeta(2k)-1) = \tfrac{3}{4}$ holds (use the definition and switch up the order of summation). So the convergence speed must be higher than that of $\tfrac{1}{k}\downarrow 0$.

The software Mathematica even evaluates the sum $\sum_{k=1}^\infty k^2(\zeta(2k)-1)$ to be $\tfrac{\pi^2}{8}$, but i don`t know how to proof this result or whether to trust it. This would mean that the true convergence speed is higher than that of $\tfrac{1}{k^3}\downarrow 0$.

Anyways, are there theorems in the literature that yield this convergence speed? Or even better: Inequalities of the form \begin{equation} \zeta(2k)-1 \leq \frac{C_\ell}{k^\ell} \qquad \text{ for } k \in \mathbb{N} \end{equation} for some explicit constant $C_\ell$ depending only on $\ell\in \mathbb{N}$? I`m not proficient in number theory and might have looked in the wrong places (such as Abramowitz Stegun so far, which only contains the series that yields the $\tfrac{3}{4}$).