Mathematical Analysis I - 1-6-7.(3)

For $n>6$ :

$\begin{aligned}\frac{5^n}{n!} &=\frac{5^n}{1\cdot 2\cdot 3\cdots n} \\ &<\frac{5^n}{(1\cdot 2\cdots 5) 6^{n-5}} \\ &=\frac{5^5}{5!} \left(\frac56 \right)^{n-5} \end{aligned}$

For any $\epsilon>0$ , we can choose $n$ large enough so that the last expression is smaller than $\epsilon$ ; hence the limit is zero .

We can use Stirling's formula to solve the problem.

$\begin{aligned} L & = \lim_{n \to \infty} \frac {5^n}\blue{n!} & \small \blue{\text{By Stirling's formula: }n! \sim \sqrt{2\pi n}\left(\frac ne \right)^n} \\ & = \lim_{n \to \infty} \frac {5^n}\blue{\sqrt{2\pi n}\left(\frac ne \right)^n} \\ & = \lim_{n \to \infty} \frac 1{\sqrt{2\pi n}}\left(\frac {5e}n \right)^n \\ & = \boxed 0 \end{aligned}$