While reading this post, I stumbled across these definitions (Wiki_german)

$$e = \lim_{n \to \infty} \sqrt[n]{n\#}$$


$$e = \lim_{n \to \infty} (\sqrt[n]{n})^{\pi(n)}.$$

The last one seems interesting, since $ \lim_{n \to \infty} (\sqrt[n]{n})=1$, proven here.

How to prove these?

  • 2
    $\begingroup$ In the second one, you can use that $\pi(n)\sim n/\log(n)$ asymptotically. $\endgroup$ – Raskolnikov Jan 5 '12 at 10:11
  • $\begingroup$ Regarding your comment, note that $\pi(n)\to\infty$, so in order for $\sqrt[n]{n}^{\pi(n)}$ to have a finite limit it is necessary for $\sqrt[n]{n}$ to converge to $1$. $\endgroup$ – Jonas Meyer Jan 5 '12 at 10:13
  • $\begingroup$ @Raskolnikov: So infact one doesn't really need $\pi(n)$ to prove it? $\endgroup$ – draks ... Jan 5 '12 at 13:39
  • $\begingroup$ Let's just say that it is an awkward way to rephrase a property of the prime-counting function. Same is probably true for the primoral $n\#$. $\endgroup$ – Raskolnikov Jan 5 '12 at 13:43
  • $\begingroup$ Definitions? Of what are those two formulas "definitions"? Anyway, the product of the primes up to $n$ is asymptotic to $e^n$ - I believe this is at the same level of difficulty as the Prime Number Theorem - that should get you the first equation. $\endgroup$ – Gerry Myerson Mar 26 '12 at 12:10

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