Question

Question #05e58

Answer 1

See below.

Explanation:

According to Stirling's asymptotic formula

`n! approx sqrt(2pi n)(n/e)^n` we have

`lim_(n->oo)(n!)^(1/n)/n = lim_(n->oo)(sqrt(2pi n)(n/e)^n)^(1/n)/n=`

`lim_(n->oo)(sqrt(2pi n))^(1/n)(n/e)/n =1 xx e^(-1) = e^(-1)`

Answer 2

First, develop your expression:

`(n!)^(1/n)/n = exp(1/nlnn!)/n`

Then you have to use a equivalence, I hope you know it
(my notations are the french ones, I don't know if you have the same):

`ln n! =_(n->+oo) nlnn -n +o(n)`

(`o(n)` means a function that is negligible when n approaches `+oo`)

You re-inject in your expression:

`(n!)^(1/n)/n =_(n->+oo) exp(1/n(nlnn -n +o(n)))/n`

`=_(n->+oo) exp(lnn -1 +o(1))/n`

`=_(n->+oo) (e^(lnn)e^ -1e^(o(1)))/n`

`=_(n->+oo) canceln/canceln e^ -1e^(o(1))`

Finally, because `o(1)` tends to `0`: `e^(o(1))` tends to `1` and

`lim_(n->+oo)(n!)^(1/n)/n = 1/e`