MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

I am looking for the asymptotic growth of product of consecutive primes. Is there anything that is known about this growth?

share|cite|improve this question
up vote 4 down vote accepted

Denote by $$\Pi(x)=\prod_{p\leqslant x}p,$$ thus $$\log\Pi(x)=\sum_{p\leqslant x}\log p:=\theta(x)\sim x,$$ which is known as the Prime Number Theorem. You may find further information in

share|cite|improve this answer

I think you're asking about the primorial function $n\sharp$, the product of all the primes less than or equal to $n$. This satisfies $ n \sharp = \exp( n(1+o(1)) ) $.

share|cite|improve this answer
Are sharper bounds known? This is actually a pretty bad estimate for n small (less than, say, a billion). – Charles Nov 1 '10 at 16:21
@Charles: If you assume the Riemann Hypothesis, you get roughly $\exp(n+O(\sqrt{n}))$ – S. Carnahan Nov 2 '10 at 3:20
@Charles: Why do you say it is bad? Up to the (very) small bound of 350000 I find results such as [ 82619, .99728] [119549, .99537] [155893, .99855] [302831, .99671] [338477, .99898] meaning that at the prime n=82619 it is about $\exp( 0.99728n )$. etc. After a relatively large prime gap the exponent will be lower and after a relatively prime dense interval it will be higher. – Aaron Meyerowitz Nov 2 '10 at 4:35
@Aaron Meyerowitz: exp(1e6)/1e6# is about 2e658. It's not so much that I mind being off by a factor of googol^6, but I wanted to know if more asymptotic terms were known. – Charles Nov 2 '10 at 5:07
@Charles OK so you are telling me that $10^6 \sharp$ is about $exp(10^6\cdot 0.99934)$. Working from that, I then get [1090697, .9991] ,[1195247, 1.00026] and [1243337, .99948]. I can't vouch for all those decimal places but the relative fluctuations should be pretty accurate. I can't imagine that asymptotic terms are going to account for fluctuations like that. The best you should expect is to get $\ln(n \sharp)$ about right and I think the answers above do that. – Aaron Meyerowitz Nov 2 '10 at 7:24

You can also prove that $$\displaystyle\lim_n\left(\prod_1^n p_i\right)^{1/p_n} = e$$

(where $p_i$ is the $i$-th prime number and $e$ is Euler's exponential number)

share|cite|improve this answer

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.