Question

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

Answer 1

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

Explanation:

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`