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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!

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2 Answers 2

up vote 3 down vote accepted

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 $g(n) = n$.

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And the preceding line may be true, I infer? –  daniel Nov 5 '12 at 22:22
    
@daniel: Dear daniel, Firstly, I fixed a typo. Secondly, yes, the statetment that $\lim_{n \to \infty} \vartheta(n)/n = 1$ is one form of the prime number theorem. Regards, –  Matt E Nov 5 '12 at 22:24
    
Yes yes. Thank you! –  daniel Nov 5 '12 at 22:25

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 $${\left( {\prod\limits_{p \leqslant x} p } \right)^{1/x}} \sim e$$

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1  
Why would one interpret $\prod_{p\le n}p$ to mean the $n$th primorial $\prod_{p\le p_n} p$? –  Erick Wong Nov 5 '12 at 22:27
    
I didn't mark you down but the statement (1) is correct as written. –  daniel Nov 5 '12 at 22:27
    
$\prod\nolimits_{p \leqslant n} p $ means the product of all primes less than or equal to $n$, which is the primorial when $n$ is prime. –  glebovg Nov 5 '12 at 22:29
    
@glebovg It is a primorial, not the $n$th one. Think $n=3$ versus $n=4$. –  Erick Wong Nov 5 '12 at 22:29
    
The OP did not specify what $n$ is, so I thought it was prime, hence the word IF in my answer. –  glebovg Nov 5 '12 at 22:31

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