The geometric mean of primes less than or equal to $x$ I want to show that the limit of the geometric mean of primes less than or equal to $x$ is $e$ as $x \to \infty$. Is this correct?
Using the product law of logarithms we have
$$\ln \prod\limits_{p \leqslant x} p  = \sum\limits_{p \leqslant x} {\ln (p)}$$
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
$${\left( {\prod\limits_{p \leqslant x} p } \right)^{1/x}} \sim e.$$
Edit: I realized that this product does not represent the geometric mean. What does it represent?
 A: The geometric mean of the collection of primes numbers less than $x$ can be written:
$$G(x)=\left( \prod_{p \le x} p\right)^{\frac{1}{\pi(x)}}$$
$$\log(G(x))=\frac{1}{\pi(x)}\sum_{p \le x}\log(p)=\frac{\vartheta(x)}{\pi(x)}$$
For large $x$, we can use the PNT and an asymptotic for the first Chebyshev function given here: 
$$\log(G(x)) \sim \frac{\log(x)}{x} \cdot x= \log(x)$$
Thus, $G(x) \sim x$. Here's a plot:

A best fit line reveals $G(x) \approx .31473x$. So I think the more interesting question to ask is what is $\displaystyle \lim_{x \rightarrow \infty} \frac{G(x)}{x}$? Can it be expressed in terms of known constants?
A: I am surprised the answer to this query is no where on the web?
Taking logs, you want to take $S(N)=\sum_{p\le N}\log p$ [the product] and then quotient this by $T(N)=\sum_{p\le N} 1$ [the number of primes] before reexponentiating for the final result. By partial summation, we have
$$S(N)=T(N)\log N-\int_2^N T(u){du\over u}$$
and the integral is $N/\log(N)$ to first order, for $T(u)\sim u/\log(u)$ by PNT. So ${S(N)\over T(N)}$ is $(\log N)-1$, and exponentiating gets you $N/e$ as the answer.
