How you solve this ? $lim_(n->oo)(5^(n)n!)/(2^(n)n^n)$

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Answer 1 · 2017-03-20T23:23:03

$lim_(n->oo)(5^(n)n!)/(2^(n)n^n)=0$

Using Stirling assymptotic approximation

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

$(n!)/n^n approx sqrt(2pi n)e^(-n)$ so

$lim_(n->oo)(5^(n)n!)/(2^(n)n^n)=(5/2)^n sqrt(2pi n)e^(-n) = (5/(2e))^n sqrt(2pi n) = a^n sqrt(2pi n)$ with $a < 1$

then

$lim_(n->oo)(5^(n)n!)/(2^(n)n^n)=0$

How you solve this ?lim_(n->oo)(5^(n)n!)/(2^(n)n^n)lim_(n->oo)(5^(n)n!)/(2^(n)n^n)