Question #05e58

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Answer 1 · 2017-05-15T19:09:22

See below.

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 · 2017-05-15T19:20:55

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

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