4
$\begingroup$

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 $\tfrac{3}{4}$).

$\endgroup$
4
  • 4
    $\begingroup$ You have $\zeta(k)=1+1/2^k+O(1/3^k)$ as $k\rightarrow\infty$. $\endgroup$
    – user334725
    Dec 15, 2021 at 21:05
  • 1
    $\begingroup$ As you probably know, one can calculate $\zeta(2k)$ explicitly for integer $k$. $\endgroup$ Dec 15, 2021 at 21:23
  • 1
    $\begingroup$ @LoïcTeyssier ... but how well can we estimate the asymptotic size of the Bernoulli numbers? $\endgroup$ Dec 16, 2021 at 0:33
  • $\begingroup$ Bernoulli numbers have the well-known asymptotics $|B_{2n}|\sim 4\sqrt{\pi n}\left( \frac{n}{\pi e}\right)^{2n}$ as $n\to\infty$. $\endgroup$ Dec 16, 2021 at 17:04

1 Answer 1

13
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ This is neat. For $N=2$ and $s = 2k$ I get e.g. $\zeta(2k) < (\tfrac{1}{2})^{2k}(1+\tfrac{2}{2k-1}) \leq 3 (\tfrac{1}{2})^{2k}$ for $k \in \mathbb{N}$ which is better than what i asked for. Thanks! $\endgroup$
    – Iceman
    Dec 15, 2021 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.