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It has been mentioned in this question Estimate for the product of primes less than n that:

In this paper Erdős shows a shorter proof for one of his old results stating that $$ s(n) = \prod_{p < n} p < 4^n$$ where the product is taken over all primes less than $n$. He also remarks that using the prime number theorem one can show $$ s(n)^{\frac1n} \stackrel{n\to\infty}{\longrightarrow} e.$$

Can someone here prove this result? It does not seem straightforward to me.

There was an answer https://math.stackexchange.com/a/204909/26632 but it's more the idea than a rigorous proof.

So I'm posting the question again, how to prove it? Best if it can be done in an elementary way.

athos
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  • Start with $\prod_{n<p\le 2n}p \le {2n \choose n} = \frac{2n (2n-1)}{n^2} {2(n-1) \choose (n-1)}\le 4 {2(n-1) \choose (n-1)}\le 4^n$ and $\prod_{p\le 4^x}p =\prod_{m\ge 1}(\prod_{4^{x/2^m}<p\le 4^{x/2^{m-1}}} p)$ – reuns Oct 04 '20 at 17:01

2 Answers2

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Assuming the prime number theorem, i.e. $$\lim_{n\to\infty}\frac{\pi(n)\,\ln n}n=1,$$ the proof is elementary, using just the following simple fact (cf. Convergence of the arithmetic mean):

If $\lim_{n\to\infty}a_n=0$, then $\displaystyle\lim_{n\to\infty}\frac{\sum^n_{k=1}a_k}n=0$.

With $$\theta(n)=\sum_{p \leq n,\,p\,\text{prime}} \ln p,$$ we have to show $$\lim_{n\to\infty}\frac{\theta(n)}n=1.$$Now (via partial summation), $$\theta(n)=\sum^n_{k=1} (\pi(k)-\pi(k-1))\,\ln k=\pi(n)\,\ln(n+1)-\sum^n_{k=1}\pi(k)\,\ln\left(1+\frac1k\right)$$ Obviously, $$a_n=\pi(n)\,\ln\left(1+\frac1n\right)=n\,\ln\left(1+\frac1n\right)\cdot\frac{\pi(n)\,\ln n}n\cdot\frac1{\ln n}\to1\cdot1\cdot0=0$$as $n\to\infty$, so $$\frac{\theta(n)}n=\frac{\pi(n)\,\ln n}n\cdot\frac{\ln(n+1)}{\ln n}-\frac1n\,\sum^n_{k=1}a_k\to1\cdot1-0=1.$$

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$s(n) = \prod \limits_{p \leq n} p$ so $\lim \limits_{n \to \infty} s(n)^{\frac{1}{n}} = e^{\lim \limits_{n \to \infty} \frac{1}{n} \sum \limits_{p \leq n} \ln p}$

And we have $ \theta(n) = \sum \limits_{p \leq n} \ln p$ and by the PNT we have that $ \lim \limits_{n \to \infty} \frac{\theta(n)}{n} =1 $

So $e^{\lim \limits_{n \to \infty} \frac{1}{n} \sum \limits_{p \leq n} \ln p} = e^{\lim \limits_{n \to \infty} \frac{\theta(n)}{n}} = e^{1} = e $

Ahmad
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  • I thought Prime number theorem only states (https://en.wikipedia.org/wiki/Prime_number_theorem#Statement) $\lim {x\to \infty }{\frac {;\pi (x);}{;\left[{\frac {x}{\log(x)}}\right];}}=1$, it doesn't talk about $ \lim \limits{n \to \infty} \frac{\theta(n)}{n} $? – athos Oct 04 '20 at 20:57
  • Do you mean $\frac{1}{n} \sum \limits_{p \leq n} \ln p$? –  Oct 05 '20 at 07:32
  • @athos exactly in the same paper look 3 or 4 lines below to see this equivalent to PNT. – Ahmad Oct 05 '20 at 11:35
  • @ProfessorVector yes, I will fix it. – Ahmad Oct 05 '20 at 11:35