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JonF
Aug2-04, 05:16 PM
does...

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

If not, is it divergent?

mathwonk
Aug2-04, 05:37 PM
isnt there somethiong called stirlings formula for n! ?? Maybe you could use that and lhopital.

mathwonk
Aug2-04, 05:40 PM
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
Aug2-04, 06:07 PM
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
Aug3-04, 03:17 AM
With Stirlings approximation: N!\approx N^Ne^{-N}, you indeed get:

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