Computing the tail of the zeta function $\sum_{n>x}n^{-s}$

Question

I want to compute $$ f_s(x)=\sum_{n>x}n^{-s} $$ for some $s>1$ (in my case, $s=3$). Of course $$ f_s(x)=\zeta(s)-\sum_{n\le x}n^{-s} $$ but for $x$ large this is hard to compute. Are there good techniques for computing this value?

Answer 1

We have $$I(s,N)=\sum_{n>N}\frac{1}{n^s}=\sum_{n>N}\frac{1}{\Gamma(s)}\int_0^{\infty}x^{s-1}e^{-n x}dx=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}e^{-N x}}{e^x-1}dx.$$ This is a typical integral with large parameter which can be analyzed by steepest descent or its variants.

In particular, the leading and subleading orders are given by $$I(s,N\to\infty)=\frac{N^{1-s}}{s-1}\left[1-\frac{s-1}{2N}+O\left(\frac1{N^2}\right)\right].$$ In principle it is possible to compute asymptotic corrections of arbitrary order $N^{-m}$. Taking into account exponential corrections may be more delicate.

Answer 2

Let $x$ be a not too small integer. Then $$\int_{x+1}^\infty t^{-s}\,dt \lt \sum_{n\gt x} n^{-s}\lt \int_x^\infty t^{-s}\,dt.$$ The integrals can be evaluated explicitly. In this way one obtains a good estimate of the tail, which can be further refined.

Answer 3

The integral bound shown by André Nicolas is the most straightforward approximation.

In order to get a more tight inequality about the difference between the integral $\int_{x}^{+\infty}t^{-s}\,dt =\frac{1}{(s-1)x^{s-1}}$ and the original series, you may use the Hermite-Hadamard inequality or the Euler-MacLaurin summation formula.