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Background: I'm trying to evaluate the Rieman Zeta function $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$

I do not need a high precision, but I need to somehow estimate the error I'm making by cutting the sum of at some point.

So here my question:

Given a $\epsilon>0$ and an $N \in \mathbb N$ (e.g. $\epsilon < 2$) is there an upper bound for $$\sum_{n=N}^\infty \frac{1}{n^{1+\epsilon}} $$?

Michael Rozenberg
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flawr
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  • This way you can evaluate $\zeta(s)$ only for $Re(s) > 1$ and it converges very slowly. You should better use that $\zeta(s) = \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ valid for $Re(s )> 0$, from which you get this formula converging much faster and for every $s \ne 1$ – reuns Dec 10 '16 at 15:44

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Hint. One may apply the integral test for convergence $$ \int_N^\infty f(x)\,dx\le\sum_{n=N}^\infty f(n)\le f(N)+\int_N^\infty f(x)\,dx $$ with $$ f(x):=\frac{1}{x^{1+\epsilon}}. $$

Olivier Oloa
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  • Oh great, that looks exactly like what I was looking for, thank you very much! – flawr Dec 10 '16 at 15:19
  • @flawr You are welcome. One then gets, for $N\ge1$, $\varepsilon>0$, $$ \left|\sum_{n=N}^\infty \frac{1}{n^{1+\epsilon}}\right|\le \frac{1}{N^{1+\varepsilon}}+\frac{1}{\varepsilon N^{\varepsilon}}. $$ – Olivier Oloa Dec 10 '16 at 15:21