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.

One (crude) attempt I tried was to consider the product $$\prod_{i=2}^n \frac{i}{\log{i}} = n!\prod_{i=2}^n \frac{1}{\log{i}}$$ which I do not know how to estimate, not to mention that I would then have to argue that it is an asymptotic estimate for $s(n).$

Is there a simple way to show the result about $s(n)$ using the prime number theorem?

  • $\begingroup$ Why would you multiply the $i/\log i$ numbers? You should multiply $n/\log n$ pieces of some specific numbers (the primes) less than $n$.. $\endgroup$
    – Berci
    Sep 30 '12 at 15:41
  • $\begingroup$ Isn't this basically the same question as this one: Show that product of primes, $\prod_{k=1}^{\pi(n)} p_k < 4^n$? $\endgroup$ Sep 30 '12 at 15:55
  • $\begingroup$ @MartinSleziak No this is not the same question. $\endgroup$
    – Jernej
    Sep 30 '12 at 19:15
  • $\begingroup$ ok. Nevertheless, it is useful to have a link to the question with a different proof of the same results here in the comments. $\endgroup$ Sep 30 '12 at 19:19
  • 3
    $\begingroup$ What Erdos says is that (product p)^(1/n)~ e, which does not imply your second line. You can't "raise both sides to the nth power" in this case. $\endgroup$
    – daniel
    Oct 5 '12 at 10:04

The reason the sum

$$ \sum_{k = 2}^{n} \frac{k}{\log k} $$

works as an estimate of the sum of all primes up to $n$ is because, roughly speaking, one on $\log N$ numbers of size around $N$ are prime. You are estimating

The sum of all primes of a given size

with the approximation

The sum of all numbers of that size, multiplied by (an estimate of) the proportion of them that are primes

(note that this method relies on the fact that the average of the primes of size around $N$ is roughly the same as the average of all numbers of size around $N$... specifically, that average is around $N$)

The analogous method for products is not dividing out by $\log i$: it is taking the $\log i$-th root: you meant to consider

$$ \prod_{k=2}^{n} k^{1 / \log k} $$

Of course, this isn't necessarily any easier to deal with. The thing to do is the one that is usually useful for products: take the logarithm. Consider

$$ \log \prod_{\substack{p=2 \\ p \text{ prime}}}^N p$$

  • $\begingroup$ Would you please consider editing your answer by replacing your $i$ by say $k$ or similar. Let me mention that to many of us $i^2=-1$ on many occasions, especially in analytical number theory. :-) $\endgroup$ Mar 28 '14 at 3:47
  • $\begingroup$ Sorry, I should have asked @Jernej. $\endgroup$ Mar 28 '14 at 3:55

The summation formula at the top of Answer 1 isn't good necessarily percentage-wise.

For instance, if n = 14, the formula gives the estimate for the sum at about 50.6, but the sum of all of the primes from 2 to 13, inclusive, is actually 41.

  • 1
    $\begingroup$ That is true, but by the prime number theorem it gets increasingly better percentage-wise for larger $n$, so even though it is $25\%$ off at $n=14$ there is some $n$ beyond which it is never more than $10\%$ off, $1\%$ off, etc. $\endgroup$ Dec 15 '15 at 6:41

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