What is $\zeta(n)$ as $n$ tends to $\infty$? How fast it goes to the limit?

Question

What is $\zeta(n)$ as $n\to\infty$? How fast it goes to the limit?

Answer 1

Absolute convergence in the right half-plane (standard for all Dirichlet series, but if you're unsure, you can compare via

$$\zeta(\sigma)<\zeta(2)\quad\forall \sigma >2$$

by just comparing term-by-term)

Then you take the limit inside, giving you

$$\lim_{\sigma\to\infty}\zeta(\sigma)=\sum_{n=1}^\infty \left(\lim_{\sigma\to\infty}n^{-\sigma}\right)=1+0+0+\ldots$$

In particular

it's easy to see by the same absolute convergence that

$$\zeta(\sigma)-1={2^\sigma\over 2^\sigma}\sum_{n=2}^\infty n^{-\sigma}=2^{-\sigma}\sum_{n=2}^\infty\left({2\over n}\right)^\sigma\sim 2^{-\sigma}$$

the last asymptotic comes from

$$\lim_{\sigma\to\infty}\sum_{n=2}^\infty\left({2\over n}\right)^\sigma =\sum_{n=2}^\infty \lim_{\sigma\to\infty}\left({2\over n}\right)^\sigma = 1+0+0+\ldots$$

from the same argument as we used on $\zeta(\sigma)\stackrel{\sigma\to\infty}{\longrightarrow}1$, where we take the limit inside and the first term is $1$, and each subsequent term goes to $0$.