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The primorial $p_n\# $ is defined as the product of the first $n$ primes: $$p_n\# = \prod_{k = 1}^n p_k.$$ Asymptotically, primorials grow like $$p_n\# = e^{(1 + o(1))n\ln n)}.$$ How does one derive this asymptotic formula?

glebovg
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    I believe this is equivalent to the Prime Number Theorem. If you want a proof of the Prime Number Theorem, this website is too small to contain one. Would you settle for a proof from the Prime Number Theorem? Also, have you looked into that book I mentioned in the comments on your earlier question? – Gerry Myerson Nov 18 '12 at 00:32
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    @GerryMyerson Again, this is not the answer I am looking for. Apostol? I will. I believe telling someone to read a book is utterly pointless, because any question can be answers in this manner and I do not think people come here to learn about good books. They come here for answers. – glebovg Nov 18 '12 at 01:09
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    Couldn't you have just said, "Yes, I'll settle for a proof from the Prime Number Theorem" (which is what you have done, by accepting the answer by @Jeff)? Don't worry about it, glebovg. You have convinced me not to try to help you. Ever. – Gerry Myerson Nov 18 '12 at 04:39
  • I do not see your answer anywhere. There is only one answer, so I can only accept that answer. I was looking for a derivation, not some random claim. Yes, the derivation employs PNT, but the statement about primorials is not equivalent to PNT. – glebovg Nov 18 '12 at 05:07
  • Did you not see that I asked you, "Would you settle for a proof from the Prime Number Theorem?"? – Gerry Myerson Nov 18 '12 at 05:57
  • I think I misinterpreted your comment. I did not realize you were offering to post an answer. I am sorry. You may post a similar, but perhaps easier derivation and I will accept your answer instead. – glebovg Nov 18 '12 at 20:14
  • I don't think I can improve on the answer posted by @Jeff. – Gerry Myerson Nov 18 '12 at 22:33
  • OK. In any case, thank you. – glebovg Nov 18 '12 at 22:47
  • @GerryMyerson Is there a proof not using PNT? – anonymous Apr 12 '25 at 07:38
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    @anon, I'm not sure. I think the result is equivalent to PNT. If that's right, it means there won't be a proof that's any easier than one using PNT. – Gerry Myerson Apr 12 '25 at 12:30
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    It might be worth having a look at https://arxiv.org/abs/2301.03586 – Gerry Myerson Apr 12 '25 at 12:39
  • @GerryMyerson: "lso, have you looked into that book I mentioned in the comments on your earlier question?" What is the that book? – pie Jun 30 '25 at 06:56
  • @pie, you're asking me about a book I mentioned in a comment on a different question in the year 2012? – Gerry Myerson Jun 30 '25 at 13:07
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    @GerryMyerson I expected that you wouldn't remember that but it was still worth trying :). – pie Jun 30 '25 at 22:58

1 Answers1

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Let $\pi(x)$ denote the prime counting function. Let $x\geq 1$. We will compute $\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)$ where the sum is taken over all primes $p\leq x$. Note that

$\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)=\displaystyle\sum_{p\leq x\;prime}\int^x_p 1/t dt=\int_1^x \frac{\pi(t)}{t}dt$.

Since there exists a constant $c$ such that $\pi(t)\leq \frac{ct}{\ln t}$, it follows that

$\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)=O(\int_1^x \frac{1}{\ln t} dt)=o(x)$.

The prime number theorem gives

$\displaystyle\sum_{p\leq x\;prime}\ln p=\pi(x)\ln x - \displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)= \ln x (\frac{x}{ln x}+o(\frac{x}{ln x}))-o(x)=x+o(x)$.

Then $\displaystyle\sum_{k=1}^n \ln p_k = \displaystyle\sum_{p\leq p_n\;prime}\ln p=p_n+o(p_n)$. It is known that $p_n=n\ln n +o(n\ln n)$; see for example the Wikipedia article on the prime number theorem. Hence $\displaystyle\sum_{k=1}^n \ln p_k =n\ln n + o(n\ln n)$. Taking exponentials of both sides gives the result you asked for.

Jeff
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