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Suppose $P(n)$ is the product of first n prime numbers, and $S(n)$ is the summation of the first n prime numbers. Does $P(n)$ grow exponentially with respect to $S(n)$, i.e., $P(n)$ can be written in the form of $e^{S(n)}$ ?

According to [1] (Sum of the first n Prime numbers), $S(n) \approx \frac{n^2}{log n}$.

Following [2] (https://en.wikipedia.org/wiki/Primorial), $P(n)\approx e^{nlog n}$.

Then $\frac{P(n)}{S(n)}\approx \frac{e^{nlogn}}{n^2 log n}$. It has exponential component, but I am not sure whether it is appropriate to say this is exponential growth.

firefighter
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2 Answers2

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$P(n) = \prod \limits_{ j \leq n} p_j = e^{\sum \limits_{j\leq n} \ln p_j} = e^{\theta(p_n)} \approx e^{p_n} \approx e^{n \ln n} = n e^{n} $

$S(n) = \sum \limits_{j\leq n} p_j \approx \int \limits_{1}^{n} p_j dj \approx \int \limits_{1}^{n} j \ln j dj = 0.5 n^2 \ln n -0.25 n^2 +0.25$

So not quite.

Ahmad
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  • According to [1] (https://math.stackexchange.com/questions/439289/sum-of-the-first-n-prime-numbers), $S(n) \approx \frac{n^2}{log n}$. Following [2] (https://en.wikipedia.org/wiki/Primorial), $P(n)\approx e^{nlog n}$. Then $\frac{P(n)}{S(n)}\approx \frac{e^{nlogn}}{n^2 log n}$. It has exponential component, but I am not sure whether it is appropriate to say this is exponential growth. – firefighter Oct 12 '20 at 23:00
  • @firefighter true but that for $n$, here its actually $p_n$ the prime in the $n$-th positions, secondly its not true that $ P(n) = e^{S(n)}$, it is sub-exponential growth – Ahmad Oct 13 '20 at 20:29
  • $e^{n\ln n}=n^n$. – Gary Sep 17 '21 at 17:04
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I think we can show that $P(n) = \exp((1+o(1))\sqrt{S(n)\ln(S(n))})$. Note that, by Landau's function, we get this immediately as an asymptotic upper bound, as $p_1 + \ldots + p_n$ is one specific partition of $S(n)$.

Let us evaluate the limit of $$ \frac{S(n)\ln(S(n))}{n^2\ln(n)\ln(n)}. $$ Using $S(n)\approx \frac{1}{2}n^2 \ln(n)$ (see other post here or here), we find that this is asymptotically equivalent to $$ \frac{0.5n^2\ln(n)(2\ln(n) + \ln(\ln(n)) - \ln(2))}{n^2\ln(n)\ln(n)}. $$ This equals $$ 1 + \frac{\ln(\ln(n))}{n^2\ln(n)\ln(n)} - \frac{\ln(2)}{n^2\ln(n)\ln(n)}, $$ which goes to one as $n$ goes to infinity. So, using the continuity of the square root function and $x \mapsto 1/x$ (which both have the fixed point $1$), we find $$ \lim_{n\to \infty} \frac{n\ln(n)}{\sqrt{S(n)\ln(S(n))}} = 1. $$ Finally, with $P(n) \approx \exp(n\ln(n))$ this gives $P(n) = \exp((1+o(1))\sqrt{S(n)\ln(S(n))})$.

Disclaimer: I am not a number theorist and only put this together by comparing various formulae I found (I was also thinking about a problem where this kind of arithmatic came up). So, please correct me if anything is wrong, or some steps are not sufficiently well explained.

Remark: In various posts here on mathstackexchange and on mathoverflow, it seems to me quite difficult to untangle if the sum of the first $n$ prime numbers or the sum of all prime numbers $\le n$ is meant, which give two different functions.

StefanH
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