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