Radical of Infinite Product of Primes I somehow stumbled across this thing.
$$\lim_{n \to \infty}({p_1\times p_2\times p_3\times\cdots \times p_n})^{\frac{1}{p_n}}$$
Where $p_i$ is the $i$-th prime.
I wanted to know if this converges or not. So I wrote a python program, and computed it up to $n = 300000$ and got this:
$2.7168952344276653$
, with $p_n = 4256249$
It seems to approach $e$, but that doesn't make sense, $e$ doesn't have anything to do with prime numbers.
It was converging extremely slowly, and the program was also very slow, so I wasn't able to compute it for higher values of $n$.
I want to know what this limit approach?
 A: Let $x$ a real number. We have, by the Abel's summation formula,
$$\sum_{p\le x}\log p = \pi(x)\log x-\int_2^x\frac{\pi(t)}{t}\,dt$$
By the Prime Number Theorem, $\pi(x)\log x \sim x$ and $\displaystyle\frac{\pi(t)}{t} =O\left(\frac{1}{\log t}\right)$ which means that $\displaystyle\int_2^x\frac{\pi(t)}{t}\,dt = O\left(\int_2^x \frac{1}{\log t}\,dt\right) = O\left(\frac{x}{\log x}\right)$
This means that $$\frac{1}{x}\sum_{p\le x}\log p =1+o(1)$$
Taking exponentials we see that the limit you're approximating is $e$.
A: Instead of
$$f(n) = \prod_{k=1}^n p_k^{1/p_n} \tag1$$
let's have a look at the log of $f$, which turns the product into a sum:
$$\ln f(x) = \frac1{p_n} \sum_{k=1}^n \ln p_k \tag 2$$
Using the approximation of the prime-counting function $\pi(x)\sim \dfrac x{\ln x}$, it's inverse is approximately (thanks to jjmath for that hint):
$$\pi^{-1}(x) \approx x\ln x \tag 3$$
Plugging this into (2):
$$\begin{align}
\ln f(x) &\stackrel{(3)}\approx \frac1{n\ln n} \sum_{k=1}^n \ln(k\ln k) \\
&= \frac1{n\ln n} \sum_{k=1}^n (\ln k +\ln\ln k) \\
&= \frac1{n\ln n} \sum_{k=1}^n (\ln (k/n) +\ln n+\ln\ln k) \tag 4\\
&= \frac1{n\ln n} \left(n\ln n+\sum_{k=1}^n (\ln (k/n) +\ln\ln k)\right) \\
&= 1 + \underbrace{\frac1{n\ln n} \sum_{k=1}^n \ln (k/n)}_{\textstyle =:A_n} + \underbrace{\frac1{n\ln n} \sum_{k=1}^n \ln\ln k }_{\textstyle =:B_n}\\
&\to 1
\end{align}$$
Up to (4) it's just using laws of logarithms, and from (4) on is just multiplying out the sum and re-arranging terms.  Both $A_n$ and $B_n$ tend to 0:
$$\begin{align}
A_n \approx \frac1{\ln n} \int_0^1 \ln x\,dx \to 0
\end{align}$$
$$\begin{align}
B_n &= \frac1{n\ln n} \sum_{k=1}^n \ln\ln k \\
&\leqslant \frac1{n\ln n} n \ln\ln n =\frac{\ln\ln n}{\ln n} \to 0\\
\end{align}$$
