Does the sequence $a^n/n!$ converge? The sequence when plotted converges to zero because a factorial grows faster than the numerator, but I can not prove that this sequence actually converges.
 A: Let $x_n = \frac {a^n} {n!}$, then
$$x_{n+1} = x_n \frac a {n+1}$$
Therefore, after some $n$, it is diminishing, and yet restricted by zero from below, so the limit exists, call it $A$.
$$\lim x_{n} = \lim x_{n+1} = A$$
$$A = \lim x_{n+1} = \lim x_n \frac a {n+1} =
\lim x_n \lim \frac a {n+1} = A \cdot 0$$
$$A = A \cdot 0 \implies A = 0$$
A: Let $x_n = {a^n \over n!}$ and note that ${x_{n+1} \over x_n} = {a \over n+1}$.
Choose $N \ge 2|a|$, then $|{x_{n+1} \over x_n}|  \le {1 \over 2}$ for $n \ge N$ and
so 
$|x_{k+N}| \le |x_N| {1 \over 2^k}$ for $k \ge 0$ and so $x_n \to 0$.
A: Use a comparison:
For some $N>0$, $n\ge N\implies n>a$. So, 
$$\frac{a^n}{n!}=\frac a1\frac a2\cdots \frac aN\frac a{N+1}\cdots\frac an.$$
If we let
$$C=\frac a1\frac a2\cdots\frac aN,$$
then
$$\frac{a^n}{n!}<C\frac an.$$
What can you say about the sequence $\frac {Ca}n$?
A: METHODOLOGY $1$:
Here is another way forward that is very efficient.  
We know from the $n$'th Term Test that if a series converges, then its terms must approach zero.
Inasmuch as the series 
$$\sum_{n=0}^\infty \frac{a^n}{n!}=e^a$$
converges, then the $n$'th Term Test guarantees that $\lim_{n\to \infty}\frac{a^n}{n!}=0$.  And we are done!

METHODOLOGY $2$:
It is easy to show that $n!>(n/2)^{n/2}$.  Therefore, we have
$$\left|\frac{a^n}{n!}\right|\le \left(\frac{\sqrt 2 |a|}{\sqrt n}\right)^n\to 0\,\,\text{as}\,\,n\to \infty$$
A: Note that the proposed solution by Andrew assumes $a\ge0$. If $a$ can be negative, then absolute values are needed, as evident by other proposed solutions. In Andrew's solution, $x_{n+1}=x_n\frac{a}{n+1}\Rightarrow \lvert x_{n+1}\rvert=\lvert x_n\rvert\lvert \frac{a}{n+1}\rvert$. Then by the same steps this leads to $\lim \lvert x_n\rvert=0$ and consequently, $\lim x_n=0$.
