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 prove 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 and Stegun so far, which only contains the series that yields the $\tfrac{3}{4}$).



Speed of convergence of $\zeta(2k)\to 1$?
Here is an explicit bound. The sum $\sum\_{n > N} n^{-s}$ for real $s > 1$ is bounded by the integral


$$\int\_N^\infty x^{-s} = N^{1-s} / (s-1).$$


Therefore for any $N$ you have


$$0 < \zeta(s) - (1 + 2^{-s} + \cdots + N^{-s}) < N^{1-s} / (s-1).$$


E.g., with $N = 3$ you get


$$0 < \zeta(s) - 1 - 2^{-s} - 3^{-s} < 3^{1-s}/(s-1).$$


