TITLE: Show that the sequence $\left(\frac{2^n}{n!}\right)$ has a limit.
QUESTION [5 upvotes]: 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
$$a_{n+1} = \frac{2^{n+1}}{(n+1)!}=\frac{2}{n+1}\frac{2^{n}}{n!}<a_n$$
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.

REPLY [4 votes]: Since $$ e^x = \sum_{r=0}^{\infty} \frac{x^r}{r!}$$ for all $x \in \mathbb{R}$, by Cauchy criterion $$\lim_{n \to \infty} \frac{x^n}{n!} = 0    $$