Proving that $\lim\limits_{n \to \infty}\frac{5^n}{n!} = 0$

Question

How does one prove this limit? $$\lim\limits_{n \to \infty}\dfrac{5^n}{n!} = 0$$ L-Hospital would work, but only in an "intuitive" sense. (Or at least, I don't see how L-Hospital would work.) I was considering $\delta$-$\epsilon$ as well, but finding a way to make, given $\epsilon > 0$, a $K(\epsilon) > 0$ such that $\forall n > K(\epsilon)$ $$\left|\dfrac{5^n}{n!}\right| < \epsilon$$ seemed to be rather difficult.

What would you suggest? Assume no further background than an introductory real analysis background (no farther than series of functions).

Answer 1

Well, so "no farther than series of functions" means we already have power and Taylor series, so

$$\sum_{n=0}^\infty\frac{5^n}{n!}=e^5\implies\;\text{the series converges}\;\implies \lim_{n\to\infty}\frac{5^n}{n!}=0$$

Answer 2

Hint: Let $a_n = \frac{5^n}{n!}$.

$$\lim_{n\to\infty}\Bigg|\frac{a_{n+1}}{a_n}\Bigg|=\lim_{n\to\infty}\frac{5^{n+1}}{(n+1)!}\frac{n!}{5^n} = \lim_{n\to\infty}\frac{5}{n+1}= 0 < 1$$

Answer 3

Then $\lim\limits_{n}\frac{5^n}{n!}\leq \lim\limits_n\frac{5^n}{10^n} =\lim\limits_n\left(\frac1{2}\right)^n = 0.$

which gives the required result by the sandwich theorem.

You just need to prove that $10^n < n!$ for all $n > N$ where you have to find $N$.

HINT: Use the representation $\frac{5^n}{n!}=\frac{5}{n}\frac{5}{n-1}\dots\frac{5}{1}$ and induction.