Why is $\lim\limits_{n\to\infty}\frac{3e^n}{n!}=0$?
$e^n=e\cdot e\cdots e$, $n$-times. I tried to find a constant $C>0$ and a $N\in \mathbb{N}$ such that $\frac{3e^n}{n!}\le C\frac{e}{n}$ for $n\ge N$, to apply the squeezing lemma afterwards. But I don't know how to estimate the fraction $\frac{3e^n}{n!}$, how can I estimate $\frac{3e^n}{n!}$ from above? Thank you.
\begin{align*} \dfrac{e^{n}}{n!}&\leq\dfrac{e}{1}\cdot\dfrac{e}{2}\cdot\dfrac{e}{3}\cdots\dfrac{e}{3},~~~~n\geq 3\\ &=\dfrac{e^{2}}{2}\cdot\left(\dfrac{e}{3}\right)^{n-2}\\ &\rightarrow 0 \end{align*}
Just as an alternative solution: the series $$\sum_{n\geq 0}\frac{x^n}{n!}$$converges absolutely for any $x\in\Bbb R$ (as easily seen by the ratio test). In particular, it converges when $x=e$. A necessary condition for series convergence is that the sequence being summed tends to zero as $n\to\infty$.
Hint1 the series $$\sum\frac{e^n}{n!}=e^e$$ converges
Hint2 Use Stirling formula Stirling's formula: proof?
$$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$$
First of all, it's obvious that $\forall n \in \mathbb{N}, u_n = \frac{3e^n}{n!}\geq 0$.
Now let's study its behaviour: $$\frac{u_{n+1}}{u_n}=\frac{e^{n+1}}{(n+1)!}\frac{n!}{e^n}=\frac{e^1}{n+1}$$, thus the sequence is decreasing, minored by $0$, thus its limit is $0$.
Every time you increase $n$, you multiply the previous term by $e$ and divide it by the new $n$. Obviously the factors are very soon smaller than $1$.
