$\lim_{n \to \infty}{a^n \over n!} $.

Question

$$\lim_{n \to \infty}{a^n \over n!}, $$ where $a >1$. I have to solve this using sandwich theorem, and I know this sequence tends to zero.

I found the lower bound of this sequence but what about upper bound.

$${1 \over n! }<{a^n \over n!}.$$

Answer 1

If you want to use the sandwich theorem, notice that you don't need a bound from bellow since $\frac{a^n}{n!}\geq 0$ for any $n$. For the bound from above let us work a bit. First of all let us fix some integer $m$ such that $a\leq m$, then $\forall~n\geq m$: $$\frac{a^n}{n!}=\frac{a^{m}a^{n-m}}{n!}\leq\frac{a^{m}m^{n-m}}{n!}\leq\frac{a^{m}m(m+1)\dots(n-1)}{n!}=\frac{a^{m}}{(m-1)!}\frac{1}{n}$$

And notice that, since $m$ is a fixed integer $\frac{a^{m}}{(m-1)!}$ is a constant, thus: $$\lim_{n\to\infty}\frac{a^{m}}{(m-1)!}\frac{1}{n}=0$$ and so, by the sandwich theorem we are done.

Answer 2

You can deal with any real $a$. Let $x_n=\frac{a^n}{n!}$.

• If $a=0$, then the sequence is $0$ and thus converges to $0$.
• If $a\neq 0$, then there exists $n_0\in\mathbb{N}$ such that $|a|\leq n$ for all $n>n_0$. For these $n$, you have $$0\leq |x_n|=\left|\frac{a}{n}\right|\left|\frac{a}{n-1}\right|...\left|\frac{a}{n_0+1}\right|\left|\frac{a^{n_0}}{n_0!}\right|\leq \left|\frac{a}{n}\right|\left|\frac{a^{n_0}}{n_0!}\right|.$$ Since $\frac{a}{n}\rightarrow_n 0$, you have that $x_n\rightarrow_n 0$ as well. Finally, you have $$x_n\rightarrow_n \begin{cases} 0^+ &\text{ if }a\geq 0, \\ 0 &\text{ if }a< 0.\end{cases}$$