real analysis - Show that the sequence $left(frac{2^n}{n!}right)$ has a limit.

Show that the sequence $\left(\frac{2^n}{n!}\right)$ has a limit.

I initially inferred that the question required me to use the definition of the limit of a sequence because a sequence is convergent if it has a limit $$\left|\frac{2^n}{n!} - L \right|{}{}<{}{}\epsilon$$

I've come across approaches that use the squeeze theorem but I'm not sure whether its applicable to my question. While I have found answers on this site to similar questions containing the sequence, they all assume the limit is $0$.

I think I need to show $a_n \geq a_{n+1},\forall n \geq 1$, so

$$a_{n+1} = \frac{2^{n+1}}{(n+1)!}=\frac{2}{n+1}\frac{2^{n}}{n!}

A monotonic decreasing sequence is convergent and this particular sequence is bounded below by zero since its terms are postive. I'm not sure whether or not I need to do more to answer the question.

Answer

It is easy to prove that
$$\sum_{k=1}^{\infty}\frac{2^n}{n!} \lt \infty$$
e.g. with the ratio test you have
$$\frac{a_{n+1}}{a_n}=\frac{n!}{2^n}\cdot\frac{2^{n+1}}{(n+1)!}=\frac{2}{n+1}\longrightarrow 0$$
Then $\lim\limits_{n\rightarrow\infty}\frac{2^n}{n!}$ has to be $0$

---

If you do not know anything about series you might assert, that $n!>3^n$ if $n\gt N_0$ for some fixed $N_0\in\mathbb{N}$. Therefore you have for those $n$
$$0<\frac{2^n}{n!} \lt\frac{2^n}{3^n}=\left(\frac{2}{3}\right)^n\longrightarrow 0$$
Thus $$\lim\limits_{n\longrightarrow\infty} \frac{2^n}{n!} =0 $$