Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I evaluate this function for given $s$?

$$\zeta(s) = \sum_{n=1}^\infty \frac1{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots$$

share|cite|improve this question
Do you mean numerically? – anon Aug 25 '10 at 7:42
I would suggest to tag it as special-function too. – Américo Tavares Aug 26 '10 at 22:40
up vote 13 down vote accepted

Assuming you're talking about numerical evaluation, along the critical strip $0<\Re s<1$ and "large" $\Im s$, (which is the region of interest for many) the Riemann-Siegel formula is standard; off the strip, what you can manage is a polyalgorithm.

For $\Re s\leq0$, one can use the reflection formula for $\zeta$,


so we can consider evaluation for $\Re s>0$ in what follows.

Note that if $|s|$ is "large enough" (how large is "large" depends on the computing environment you're in), one can simply use the defining series ($\sum_{j=1}^\infty \frac{1}{j^s}$), since its terms quickly diminish in magnitude. That leaves the problem of what to do for small to medium-sized $s$.

The key is to use the Dirichlet $\eta$ function:

$$\eta(s):=-\sum_{j=1}^\infty \frac{(-1)^j}{j^s}$$

which is related to $\zeta$ by the following identity:


The reason for our interest in $\eta$ is that although this sum is slowly convergent, it is an alternating series, and a number of algorithms exist for quickly finding the sum of an alternating series numerically.

One method is the Levin transformation; another one, which is one of the simplest methods for numerically summing alternating series (and my personal favorite) is the Cohen-Rodriguez Villegas-Zagier algorithm. The algorithm is a bit too long to describe here, so I will just have to point you to the original paper.

This is in fact identical to the approach taken by Borwein in this paper.

share|cite|improve this answer
In Example 3 (p.6) of the Cohen, Rodriguez, Villegas and Zagier's paper the values $\zeta (1/2)$ and $\zeta (-1+i)$ of the Riemann zeta function are calculated. "The latter works even though the coefficients $a_{k}$ of our "alternating" series $\sum (-1)^{k}a_{k}$ are not alternating or even real and do not tend to zero."! – Américo Tavares Aug 25 '10 at 9:37
+1 for the Riemann-Siegel formula and Cohen-Rodriguez Villegas-Zagier algorithm. – Américo Tavares Aug 25 '10 at 10:57
Though CRVZ was used for evaluating $\zeta$ at $i-1$ in the paper, some numerical testing I did shows that the method can be unstable for some arguments with negative real part. This is why I recommended using the reflection formula instead for arguments in that region. – J. M. Aug 25 '10 at 16:31
On another note, Maple and Mathematica employ Euler-Maclaurin summation for computing $\zeta$; the problem with this of course is that there is the assumption that either you can easily generate or have a large enough cache of Bernoulli numbers (which is not a problem for Maple and Mathematica, but may be inconvenient in other environments). – J. M. Aug 25 '10 at 16:34
Yet another note: in fact, due to $\zeta$'s mirror symmetry ($\zeta(\bar{s})=\overline{\zeta(s)}$), one can actually restrict computations so that one is handling only arguments with real and imaginary part both positive. – J. M. Aug 27 '10 at 1:29

You should look at the work of Wadim Zudilin. In particular you should look at "One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational" (Turpion link, pdf 91k, gzip ps 80k) in Russian Math. Surveys 56:4 (2001), 774--776;

share|cite|improve this answer
I add the work of Tanguy Rivoal on the same subject. For instance this (in French). – Américo Tavares Aug 25 '10 at 21:19
There are of course special methods for evaluating $\zeta$ at even (using of course the relation with the Bernoulli numbers) and odd integers. Searching the literature should turn up the methods for $\zeta(2k+1)$. – J. M. Aug 25 '10 at 23:50
@J. Mangaldan Yes, of course. Euler was the first to give the formula for $\zeta(2)$ with several proofs, and as you say $\zeta(2n)$ can be expressed using the Bernoulli numbers $B_\nu$ (was that Euler too?). If someone could add a link it would be good. – AD. Aug 26 '10 at 5:55
Indeed, it was Euler who computed values at even arguments from 2 to 26; and T.J. Stieltjes later extended these calculations by Euler. – J. M. Aug 27 '10 at 1:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.