Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why is

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

Can we generalize it to any exponent $x \in \Bbb R$? This is to say, is

$$\lim_{n \to \infty} \frac{x^n}{n!}=0\text{ ?}$$


This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions.

and here: List of abstract duplicates.

share|improve this question
6  
So $\frac{2^n}{n!}$ is always positive, right? If you can show that $\frac{2^{n+1}}{(n+1)!} \leq \frac{2^n}{n!}$ is always so, then... –  J. M. Oct 31 '11 at 16:59
2  
Thank you J.M., your solution was simple and worked well. I wish you had provided it in the form of an answer so that I could accept it! –  Matt Nashra Oct 31 '11 at 17:05
8  
Then the sequence converges, but not necessarily to zero. –  AMPerrine Oct 31 '11 at 17:06
    
Note that for $n \ge 4$, $n!=(6)(4\cdot 5\cdots n)$. But $4/2\ge 2$, $5/2 \ge 2$, and so on, so $\frac{2^n}{n!} \le \frac{8}{6}\frac{1}{2^{n-3}}$. –  André Nicolas Oct 31 '11 at 17:07
1  
@JM: $0<\frac{1}{2}+2^{-(k+1)}<\frac{1}{2}+2^{-k}$, but that sequence does not converge to $0$. –  robjohn Oct 31 '11 at 17:39
show 2 more comments

9 Answers 9

Consider that $$\frac{2^n}{n!} = \frac{\overbrace{2\times 2\times\cdots \times 2}^{n\text{ factors}}}{1\times 2 \times \cdots \times n} = \frac{2}{1}\times \frac{2}{2}\times \frac{2}{3}\times\cdots \times\frac{2}{n}.$$ Every factor except the first two is smaller than $1$, so at each step you are multiplying by smaller and smaller numbers, with the factors going to $0$.

share|improve this answer
add comment

First you show that $n!>3^n$ and then use $$ \lim\limits_{n}\frac{2^n}{n!}\leq \lim\limits_n\frac{2^n}{3^n} =\lim\limits_n\left(\frac2{3}\right)^n = 0. $$

To show that $n!>3^n$ you use induction. For $n = 7$ it holds, you assume that it holds for some $k\geq7$ then $(k+1)! = k\cdot k!>k\cdot 3^k>3^{k+1}$ since $k\geq 7>3$.

share|improve this answer
2  
you should prove that for $n=7$ it works... :P –  Valerio Capraro Oct 31 '11 at 17:55
10  
@Valerio At this stage, it is conventional to say "It can be trivially verified that the inequality holds for $n=7$." :-) –  Srivatsan Oct 31 '11 at 18:18
add comment

The series for $e^2$ $$ \sum_{k=0}^\infty\frac{2^n}{n!} $$ converges by the ratio test. The terms of a convergent series must tend to $0$.

share|improve this answer
    
Thanks to Mike Spivey for noticing my typo. –  robjohn Oct 31 '11 at 17:46
    
Would the downvoter care to comment? –  robjohn Aug 26 '13 at 10:45
add comment

I am surprised that noone mentioned this:

$$2 \cdot 2 \cdot 2... \cdot 2 \leq 2 \cdot 3 \cdot 4... \cdot (n-1)$$

Thus $2^{n-2} \leq (n-1)!$.

Hence

$$0 \leq \frac{2^n}{n!} \leq \frac{4(n-1)!}{n!}=\frac{4}{n} \,.$$

Generalization Let $x$ be any real number.

Fix an integer $k$ so that $\left| x \right| <k$.

Then, for all $n> k$ we have:

$$\left| x\right| ^{n-k} < k(k+1)(k+2)...(n-1) $$

Thus

$$0 < \frac{\left|x \right|^n}{n!} \leq \frac{\left|x\right|^kk(k+1)(k+2)...(n-1)}{n!}=\frac{\left|x \right|^k}{(k-1)!}\frac{1}{n}$$

Since $k$ is fixed, $\frac{\left|x \right|^k}{(k-1)!}$ is just a constant, thus $\lim_n \frac{\left|x \right|^k}{(k-1)!}\frac{1}{n}=0$.

By Squeeze theorem, we get that

$$\lim_n \left| \frac{x ^n}{n!} \right|= \lim_n \frac{\left|x \right|^n}{n!}=0 \,.$$

Now, since $\lim_n \left| \frac{x ^n}{n!} \right|=0$, we get

$$\lim_n \frac{x ^n}{n!} = 0\,.$$

P.S. A more general result applicable in this case is the following:

Lemma If $a_n$ is a sequence so that

$$\limsup_n |\frac{a_{n+1}}{a_n}| <1$$ then $\lim_n a_n =0$.

share|improve this answer
    
This one is my favorite because it is a very simple and elegant explanation! –  Argon Apr 20 '12 at 1:44
    
@N.S. That is a very nice idea. Would you mind generalizing it to any exponent $x$? –  Pedro Tamaroff Apr 20 '12 at 1:45
    
@PeterT.off Done –  N. S. Apr 20 '12 at 1:56
    
@N.S. Great! Now we have two proofs. Maybe someone else can complete the "Gaussian-like" triad =D. –  Pedro Tamaroff Apr 20 '12 at 1:58
add comment

Define the sequence $\{ a_n\}$ as $a_n= \dfrac{x^n}{n!}$ for $x\in \mathbb R$ and $n\in \mathbb N$.

  1. If $x=0$, it is trivial that $\lim a_n=0$

  2. If $x>0$, then one has that

    • For $n\in \Bbb N$, $a_n >0$.
    • For $n$ sufficiently large (say $n \geq x$), it will be the case $$a_{n+1} = \frac{x^{n+1}}{(n+1)!}=\frac{x}{n+1}\frac{x^{n}}{n!}<a_n.$$ This means that after certain $n$, $a_{n+1}<a_{n}$.
    • Since a bounded monotonically decreasing sequence of real numbers must have a limit, $$a= \lim_{n\to\infty} a_n=\lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty}\frac{x}{n+1}\cdot\lim_{n\to\infty} a_n = 0\cdot a$$ $$\implies a=0.$$
  3. If $x <0$, we introduce a $(-1)^n$ factor. Since we've proven that $a_n$ goes to zero, we use the property that if $\{ b_n \}$ is bounded and $a_n \to 0$, then $\lim\limits_{n\to\infty} a_n\cdot b_n =0$, and we're done.

share|improve this answer
    
Any feedback on the downvote? –  Pedro Tamaroff Apr 19 '12 at 19:04
1  
Hmm. You know, the -1 goes away if you delete the answer. If you make a complete version, you can cut and paste it into a new answer. –  Kaz Apr 19 '12 at 23:30
1  
Looking at the edit history, it might have been because you pulled $n$ outside of a $\lim\limits_{n\to\infty}$ (and forgot a $\lim$ between two of the = signs). Otherwise looked okay. –  anon Apr 19 '12 at 23:31
    
@anon I'm polishing it. I generalized something which wasn't correct. –  Pedro Tamaroff Apr 19 '12 at 23:32
2  
Yes, the original version looked fine to me except for minor issues, not worth downvoting. As to the $|a_n|$, yes, but that is trivial, isn't it? –  Aryabhata Apr 19 '12 at 23:41
show 13 more comments

This was here before. I'll recreate what I said then.

The basic idea is that $n! > (n/2)^{n/2}$ (by looking at the terms beyond $n/2$).

So $x^n/n! < x^n/(n/2)^{n/2} = (x^2)^{n/2}/(n/2)^{n/2} = (2x^2/n)^{n/2}$.

So$^2$, if $n > 4x^2$, $x^n/n! < 1/2^{n/2}$ which goes nicely to zero - about as elementary as can be.

share|improve this answer
add comment

$$\lim_{n \to \infty} \frac{2^n}{n!}=0$$ I just want to give an intuitive idea why this limit is zero. I'm not solve the problem with math . From the logical ground it can be proved.Note that as n tends to $\infty$, $2^n$ and ${n!}$ both tends to $\infty$ but ${n!}$ tends to $\infty$ more rapidly than $2^n$. This fact shows that the limiting value is $0$.

share|improve this answer
1  
That's a good idea, but how do you know that $n!$ tends to $\infty$ more rapidly than $2^n$ ? –  robjohn Oct 25 '12 at 8:01
add comment

The Stirling's formula says that:

$$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n, $$

inasmuch as

$$ \lim_{n \to \infty} \frac{n!}{\sqrt{2 \pi n} \left(\displaystyle\frac{n}{e}\right)^n} = 1, $$

thearebfore

$$ \begin{aligned} \lim_{n \to \infty} \frac{2^n}{n!} & = \lim_{n \to \infty} \frac{2^n}{\sqrt{2 \pi n} \left(\displaystyle\frac{n}{e}\right)^n} = \lim_{n \to \infty} \Bigg[\frac{1}{\sqrt{2 \pi n}} \cdot \frac{2^n}{\left(\displaystyle\frac{n}{e}\right)^n} \Bigg]\\ &= \lim_{n \to \infty} \frac{1}{\sqrt{2 \pi n}} \cdot \lim_{n \to \infty} \left(\frac{e2}{n}\right)^n = 0 \cdot 0^\infty = 0 \end{aligned} $$

Note: You can generalize replacing $2$ by $x$.

Visit: http://en.wikipedia.org/wiki/Stirling's_approximation

share|improve this answer
add comment

The simplest way would be; let $$ \color{fuchsia}{P_n=\frac{x^n}{n!}=} \color{maroon}{\frac x1.\frac x2.\frac x3\cdots\frac x{x-1}.\frac xx.\frac x{x+1}\cdots\frac x{n-1}.\frac xn}$$ Then $$\color{maroon}{0}\color{red}{<}\color{fuchsia}{P_n}\color{red}{<}\color{maroon}{\frac11.\frac12\cdots\frac{x-1}{x-1}.\frac xx.}\color{green}{\frac x{x+1}.\frac x{x+1}\cdots\frac{x}{x+1}.\frac x{x+1}}$$ Or $$\color{maroon}{0}\color{red}{<}\color{fuchsia}{P_n}\color{red}{<}\color{maroon}{\frac{x^x}{x!}.}\color{green}{\left(\frac x{x+1}\right)^{n-x}}$$ And as $$\color{fuchsia}{\lim_{n\to\infty}\color{maroon}{0}=0}\\ \color{fuchsia}{\lim_{n\to\infty}\color{maroon}{\frac{x^x}{x!}.}\color{green}{\left(\frac x{x+1}\right)^{n-x}}=0}$$ By using $\color{red}{\text{Sandwich theorem}}$ the result can be obtained; I leave you to read between the lines.

share|improve this answer
add comment

Your Answer

 
discard

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.