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?

Pretty good is to approximate the tail by an integral. One can if needed refine further, but the integral will give the main term of the error. — André Nicolas, Sep 11 '15 at 16:26
@AndréNicolas: I'm looking for, say, 20-100 significant digits with $x\approx10^{20}$ -- is there a way of refining this technique to give that kind of precision? — Charles, Sep 11 '15 at 16:36
Lots of digits! For $\zeta(3)$ the error term is of order $\frac{1}{x^2}$, so we do pretty well for the very large $x$ that you allow. For further refinement (and to cut down on the $x$) please see the Euler-Maclaurin summation formula. — André Nicolas, Sep 11 '15 at 16:43

Start wearing purple · Accepted Answer · 2015-09-11T17:12:59.030

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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.

edited Sep 11 '15 at 17:12

answered Sep 11 '15 at 16:51

Start wearing purple

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Dear Start wearing purple, can you give a reference for your statement on the leading and subleading orders? – R.J. Etienne Sep 22 '18 at 18:38

score 2 · Answer 2 · answered Sep 11 '15 at 16:39

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.

score 1 · Answer 3 · answered Sep 11 '15 at 16:48

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.

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

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