Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $n$ tending to infinity find the following limit $$2^n/n!$$ i have a feeling that it is multiplication of many numbers with the last one turning to 0 but the 1st 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 .

share|cite|improve this question
Try the ratio test. – aschepler Aug 9 '14 at 15:05
ratio test ???? – avz2611 Aug 9 '14 at 15:06
See – Paul Sundheim Aug 9 '14 at 15:08
ok got it , thank you for the help , – avz2611 Aug 9 '14 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 $0$. – Vishwa Iyer Aug 10 '14 at 0:42
up vote 22 down vote accepted

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

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

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

share|cite|improve this answer
Hey! I recognize this one!! – I like Serena Aug 9 '14 at 16:29



share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

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 $n \to \infty$

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

share|cite|improve this answer

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

share|cite|improve this answer
wow this is new to me , so i am going to like this one – avz2611 Aug 9 '14 at 15:11
You're missing a power of $n$ on the RHS. – Jam Aug 9 '14 at 15:18
Every time a factorial appears somewhere in a limit, someone recommends Stirling's approximation. – Gahawar Aug 9 '14 at 15:19
Ridiculous hint. Verifying it is significantly harder than the question at hand. – Andrés E. Caicedo Aug 11 '14 at 15:18
I don't think so, it is much simpler... – MathGems Aug 15 '14 at 12:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.