Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

We know it converges for any prime p. I just want to know how to compute its exact value: $$\prod_{n=1}^{\infty} (1-p^{-n})$$

share|cite|improve this question
I could be wrong here, but I doubt this has a closed form evaluation. Or, do you want to ask for computing approximations. In any case this is a particular value of the Euler function (or Q-Pochhammer symbol). It would also help in assesing whether this question is on-topic here, if you could give some context why and what exactly you want to know related to this. (Cf FAQ and How to ask). –  quid Aug 14 '12 at 10:57
I don't think that's right: when $p=2$ you get $0.288788\dots$. Is there an error in how the question was phrased? –  Henry Cohn Aug 14 '12 at 12:39
I'm away from my references, but I'm pretty sure it's not $1/4$ when $p=2$. It's related to the Dedekind eta-function. –  Gerry Myerson Aug 14 '12 at 12:41
Numerical evaluation suggests that this product is 0.288788, not 0.25, at $p=2$. (I computed 20 terms and then 30 terms and got the same answer to 6 digits.) It shouldn't be hard to rigorously show that the product is more than 0.26. –  David Speyer Aug 14 '12 at 12:42
This product also shows up in the Cohen-Lenstra heuristics for the distribution of p-Sylow subgroups of class groups of imaginary quadratic fields. This is expected to be related to Andreas Blass's comment about the probability that an n by n matrix over F_p is nonsingular, thanks to a paper of Friedman and Washington. –  Simon Aug 14 '12 at 16:11

2 Answers 2

up vote 3 down vote accepted

Wolfram MathWorld gives the expression of this product in terms of the q-Pochhammer symbol and the Jacobi theta function. See formulas (46) and (47) in

share|cite|improve this answer
Well, it is a q-Pochhammer symbol essentially by definition (as I commented). –  quid Aug 14 '12 at 12:35
Yes, it is. But its exact value is still very difficult. –  mason Aug 14 '12 at 14:18
Determining its exact value is unlikely to be a well-posed question. –  Jack Huizenga Aug 14 '12 at 20:16

Let's fix the issue of giving bounds on the infinite product. The pentagonal number theorem gives, after grouping pairs of consecutive terms with the same sign, an alternating series with terms that are decreasing in modulus. So for instance for $c:=\prod _ {n\ge1} (1-2^{-n})$ one has

$$1-\frac{1}{2}-\frac{1}{4}+\frac{1}{32}+\frac{1}{128} -\frac{1}{4096}-\frac{1}{ 32768 } < c < 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{32}+\frac{1}{128}$$ that is $ 0.288787842 < c < 0.2890625, $ in any case larger than $1/4$.

share|cite|improve this answer
+1 for including this information and thus avoiding the possiblilty of any confusion on this matter. –  quid Aug 14 '12 at 20:15
You can find it in GTM 245, P209 , ex 1. –  mason Aug 19 '12 at 1:41
Now even in the notes of my calculus course, why. ;-) –  Pietro Majer Aug 21 '12 at 6:18
Note that this problem has nothing to do with primes. The product converges if $|p|>1$. –  Marc Chamberland Aug 24 '12 at 17:38

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.