0
$\begingroup$

Let p be prime. Assume (1): $\hspace{10mm} (\prod_{p\leq n} p)^{1/n} \sim e.$

Then $(e^{\ln \prod p})^{\frac{1}{n}} = e^{(\sum \ln p)/n} \sim e \implies \lim_{n=1}^\infty \frac{e^{(\sum \ln p)/n}}{e} = 1. $

And so

$ \lim_{n=1}^\infty~ e^{(\frac{\sum \ln p}{n}-1)} = 1$ or

$\lim \frac{\sum \ln p}{n } - 1 = 0 \implies \lim (\frac{\vartheta(n)}{n}- 1) = 0$

But this implies that $\lim_{n = 1}^\infty~ (\vartheta(n) - n) = 0 $

which is false.

I have good reason to think (1) is true so perhaps someone can point to the error, which I will chalk up to hurricane-fatigue.

Thanks!

2 Answers 2

3

If $f(n) \to 0$ and $g(n) \to \infty,$ then a priori you can't say anything about the product $f(n)g(n)$. In your case, $f(n) = (\vartheta(n)/n) - 1$ and $g(n) = n$.

  • 0
    Yes yes. Thank you!2012-11-05
0

If $\prod\nolimits_{p \leqslant n} p $ is supposed to be the primorial ${p_n}\# $ then (1) is incorrect because primorial grows like ${e^{(1 + o(1))x\ln (x)}}$ i.e. ${p_n}\# \sim {e^{(1 + o(1))n\ln (n)}}$, so raising it to $1/n$ gives ${n^{1 + o(1)}} \ne e$.

Edit: Regarding your second question i.e. proof of (1), note that $\ln \prod\limits_{p \leqslant x} p = \sum\limits_{p \leqslant x} {\ln (p)}$ but $\sum\limits_{p \leqslant x} {\ln (p)} = \vartheta (x) \sim x$ hence $\prod\limits_{p \leqslant x} p \sim {e^x}$ and the geometric mean is ${\left( {\prod\limits_{p \leqslant x} p } \right)^{1/x}} \sim e$

  • 0
    @glebovg It's fine to assume $n$ is prime, but this is inconsistent with comparing the product of primes up to $n$ with $p_n\#$, which is the product of primes up to $p_n$.2012-11-05