Does the sequence $a^n/n!$ converge?

Question

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.

Answer 1

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 $A$.

$$\lim x_{n} = \lim x_{n+1} = A$$

$$A = \lim x_{n+1} = \lim x_n \frac a {n+1} = \lim x_n \lim \frac a {n+1} = A \cdot 0$$

$$A = A \cdot 0 \implies A = 0$$

Answer 2

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 $\frac {Ca}n$?

Answer 3

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!

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METHODOLOGY $2$:

It is easy to show that $n!>(n/2)^{n/2}$. Therefore, we have

$$\left|\frac{a^n}{n!}\right|\le \left(\frac{\sqrt 2 |a|}{\sqrt n}\right)^n\to 0\,\,\text{as}\,\,n\to \infty$$

Answer 4

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 $x_n \to 0$.

Answer 5

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, $\lim x_n=0$.