# 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.

• Try the ratio test. Aug 9, 2014 at 15:05
• ratio test ???? Aug 9, 2014 at 15:06
• Aug 9, 2014 at 15:08
• ok got it , thank you for the help , Aug 9, 2014 at 15:09
• You can see that the factorial function grows much faster than the exponential function (meaning that it increases much faster), therefore the limit will converge to $0$. Aug 10, 2014 at 0:42

$$0<\frac{2^n}{n!}=\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \dots \cdot \frac{2}{ n} \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \dots \cdot \frac{2}{3}=\frac{2}{1} \cdot \frac{2}{2} \cdot \left (\frac{2}{3} \right )^{n-2}=2 \left ( \frac{2}{3} \right )^{n-2}$$

As $n \rightarrow \infty$, $\left ( \frac{2}{3} \right )^{n-2} \rightarrow 0$

Therefore, from the Squeeze Theorem $$\lim_{n \rightarrow \infty} \frac{2^n}{n!}=0$$

• Hey! I recognize this one!! Aug 9, 2014 at 16:29

BIG HINT:

$$\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}=e^x<\infty$$

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 $$\lim_{n\to\infty }\frac{2^n}{n!}=0.$$

Hint: You can use Stirling's approximation $$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

• wow this is new to me , so i am going to like this one Aug 9, 2014 at 15:11
• You're missing a power of $n$ on the RHS.
– Jam
Aug 9, 2014 at 15:18
• Every time a factorial appears somewhere in a limit, someone recommends Stirling's approximation. Aug 9, 2014 at 15:19
• Ridiculous hint. Verifying it is significantly harder than the question at hand. Aug 11, 2014 at 15:18
• I don't think so, it is much simpler... Aug 15, 2014 at 12:16

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 $n \to \infty$

$$a = a(\lim_{n \to \infty}\frac{2}{n+1}) = 0$$

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 $n > (3a)^3/2$, $\dfrac{a^n}{n!} <\left(\dfrac{1}{2}\right)^n$.

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 $\dfrac{2^n}{n!}.$

Hint: Use induction to prove $\frac{2^n}{n!}\le (\frac{1}{2})^{n-4}$ for $n\ge 4$.

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 $$\lim_{n\to\infty} f_n=e^{-\infty}=0$$.