Proof That $\lim_{n \rightarrow \infty} \frac{n}{(n!)^\frac{1}{n}} = e$

[JonF]

does...

\lim_{n \rightarrow \infty} \frac{n}{(n!)^\frac{1}{n}} = e

If not, is it divergent?

 

[mathwonk]

isnt there somethiong called stirlings formula for n! ?? Maybe you could use that and lhopital.

 

[mathwonk]

well i just looked up stirling and it seems to suggest at a quick calculation, not guaranteed, that this limit is e/sqrt(2pi)

 

[JonF]

e/(2pi)^(1/2) aprox= 1.0844

my calc can do the limit up to 200 and it equals about 2.67021... that's why i thought it may = e

 

[Galileo]

With Stirlings approximation: N!\approx N^Ne^{-N}, you indeed get:

\frac{N}{(N^Ne^{-N})^{\frac{1}{N}}}=\frac{N}{Ne^{-1}}=e