How to calculate limit of a function having factorial in denominator For $n$ tending to infinity find the following limit
$$\frac{2^n}{n!}.$$
I have a feeling that it is multiplication of many numbers with the last one turning to $0$ but the first one is finite so limit should be $0$. But I am not sure and neither am I able to put it in mathematical form.
Thank you for your help.
 A: You can use the theorem of d'Alembert for the sequences then you immediately have:
if $x_n=\frac{2^n}{n!}$,
$$\lim_{n\to\infty }\left|\frac{x_{n+1}}{x_n} \right|=\lim_{n\to\infty }\frac{2^{n+1}n!}{(n+1)! 2^n}=\lim_{n\to\infty }\frac{2}{(n+1)}=0<1$$
then $$\lim_{n\to\infty }\frac{2^n}{n!}=0.$$
A: $$0<\frac{2^n}{n!}=\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \dots \cdot \frac{2}{ n} \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \dots \cdot \frac{2}{3}=\frac{2}{1} \cdot \frac{2}{2} \cdot \left (\frac{2}{3} \right )^{n-2}=2 \left ( \frac{2}{3} \right )^{n-2}$$
As $n \rightarrow \infty$, $\left ( \frac{2}{3} \right )^{n-2} \rightarrow 0$
Therefore, from the Squeeze Theorem $$\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0$$
A: Hint: You can use Stirling's approximation $$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n
$$
A: One way to aproach these kinds of limits is to use the monotone convergence theorem, (real bounded monotone sequences converge). So for convergence you need to prove that 1. your sequence is monotone, 2. it's bounded
For your sequence you can prove that it is decreasing by using the ratio test as in idm's answer. And you can clearly see that it is bounded by 0. This means that a limit exists, let $a_n$ be your sequence, then
$$
a_{n+1} = \frac{2^{n+1}}{(n+1)!} = a_n\frac{2}{n+1}
$$
Now because we know $\lim_{n \to \infty} a_n = a$, we can replace $a_n$ and $a_{n+1}$ in the above equation by their limit, when $n \to \infty$
$$
a = a(\lim_{n \to \infty}\frac{2}{n+1}) = 0
$$
A: For any $a > 0$,
$\lim_{n \to \infty} \dfrac{a^n}{n!}
=0$.
To see this,
note that,
$n!
\ge (2n/3)^{n/3}
$
by looking at the last
$n/3$ terms.
Therefore
$\dfrac{a^n}{n!}
\le \dfrac{a^n}{(2n/3)^{n/3}}
=\left(\dfrac{3a}{(2n)^{1/3}}\right)^n
$.
By making $n$
large enough compared with $a$,
this can be made
as small as we want.
For example,
if $n > (3a)^3/2$,
$\dfrac{a^n}{n!}
<\left(\dfrac{1}{2}\right)^n
$.
A: Alternatively, using induction, it's easy to see that $\forall \, n \in \mathbb{N}, \left(\dfrac{n}{e}\right)^n < n!$ (the inductive step uses the definition of $e$.) Then by the corollary to the Archimedean property of the  real numbers, there exists $N \in \mathbb{N}$ such that $N > 2e .$ Then $\forall \, n \geq N$ we have:
$0 < \dfrac{2^n}{n!}<\left(\dfrac{2e}{n}\right)^{n}< \left(\dfrac{2e}{N}\right)^{n}$
and since $\dfrac{2e}{N}<1,\left(\dfrac{2e}{N}\right)^{n} \to 0,$ and by the squeeze theorem so does $\dfrac{2^n}{n!}.$
A: BIG HINT:
$$\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}=e^x<\infty$$
A: Hint: Use induction to prove $\frac{2^n}{n!}\le (\frac{1}{2})^{n-4}$ for $n\ge 4$.
A: Not a mathematician, but this is another alternative. Set $f_n=\frac{2^n}{n!}$, and define the auxiliary function $g_n=\ln(f_n)$. That gives
$$ g_n=n\ln(2)-\ln(n!)=n\ln(2)-\color{red}{\bigg(\ln(n)+\ln(n-1)+\dots+\ln(2)+\ln(1)\bigg)}.$$
With every increase in $n$, the $n\ln(2)$-term grows by 0.69, while the red-colored term removes $\ln(n)$. As $\ln(n)> 0.69$ for $n>2$ and $\ln$ is a monotously increasing function, that means that from $n=3$ onwards the series decreases faster than it can possibly grow. For example:
\begin{align}
   g_1 &= \bigg(\ln(2)\bigg) - \color{red}{\bigg( 0 \bigg)}=0.69,\\\\
   g_2 &= \bigg(\ln(2)+\ln(2)\bigg)-\color{red}{\bigg(\ln(2)\bigg)}=0.69,\\\\
   g_3 &= \bigg(\ln(2)+\ln(2)+\ln(2)\bigg)-\color{red}{\bigg(\ln(3)+\ln(2)\bigg)}=0.29,\\\\
   g_4 &= \bigg(\ln(2)+\ln(2)+\ln(2)+\ln(2)\bigg)-\color{red}{\bigg(\ln(4)+\ln(3)+\ln(2)\bigg)}=-0.41,\\\\
   g_5 &= \bigg(\ln(2)+\ln(2)+\ln(2)+\ln(2)+\ln(2)\bigg)-\color{red}{\bigg(\ln(5)+\ln(4)+\ln(3)+\ln(2)\bigg)}=-1.32 \\\\
   \vdots
\end{align}
Hence, $\lim_{n\to\infty} g_n=-\infty$, which means that with $\lim_{n\to\infty} g_n=\lim_{n\to\infty}\ln(f_n)$ we obtain $\lim_{n\to\infty} f_n=e^{-\infty}=0$.
A: For sufficiently large $n$ we have:
\begin{align*}
0 < \frac{2^n}{n!} = \frac{2}{1} \times \frac{2}{2} \times \frac{2}{3} \times \cdots \times \frac{2}{n-1} \times \frac{2}{n} < \frac{2}{1} \times 1 \times 1 \times \cdots \times 1 \times \frac{2}{n} = \frac{4}{n} \stackrel{n \to \infty}{\longrightarrow} 0
\end{align*}
