# Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$.

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.

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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... –  Guess who it is. Oct 31 '11 at 16:59
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
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
@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

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 $k\geq 7>3$.

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you should prove that for $n=7$ it works... :P –  Valerio Capraro Oct 31 '11 at 17:55
@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

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 $0$.

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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 $0$.

Alternate Approach to the Second Question

Inspired by Ilya, I have moved my deleted answer from another question here.

For $n\ge2x$, we have \begin{align} \frac{x^n}{n!} &=\frac{x^{\lfloor2x\rfloor}}{\lfloor2x\rfloor!}\frac{x}{\lfloor2x+1\rfloor}\frac{x}{\lfloor2x+2\rfloor}\cdots\frac{x}{n}\\[4pt] &\le\frac{x^{\lfloor2x\rfloor}}{\lfloor2x\rfloor!}\left(\frac12\right)^{n-\lfloor2x\rfloor} \end{align} Since $$\lim_{n\to\infty}\left(\frac12\right)^{n-\lfloor2x\rfloor}=0$$ we have $$\lim_{n\to\infty}\frac{x^n}{n!}=0$$

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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
And here we go into the dark past of a person who also used series test to show convergence of elementary sequence (and even got some votes for that) :) I didn't downvote though –  Ilya Nov 28 '14 at 12:06
@Ilya: I've moved my answer from the duplicate question here since it answers the second part. –  robjohn Nov 28 '14 at 14:09
Now I can happily upvote it :) –  Ilya Nov 28 '14 at 14:11

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 $\lim_n a_n =0$.

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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

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.

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Any feedback on the downvote? –  Pedro Tamaroff Apr 19 '12 at 19:04
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
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
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

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

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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.

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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.

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

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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

Let $\:\epsilon>0$.

The fact that $x$ is fixed tells you that there exist $M\in \mathbb{N}$ such that $|x|<M$.

With this you have that there exist $N\in\mathbb{N}$ such that $\displaystyle\left(\frac{M^M}{M!}\right)\frac{1}{N}<\epsilon$

Then, if $n\geq MN$

$\displaystyle\left\|\frac{x^n}{n!}-0\right\|\leq\frac{M^n}{n!}\leq\frac{M}{1}\cdots\frac{M}{M}\cdots\frac{M}{MN}\leq\left(\frac{M^M}{M!}\right)\frac{1}{N}<\epsilon$

So $\displaystyle \lim_{n\to\infty}\frac{x^n}{n!}=0$

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$u_n=\dfrac{x^n}{n!} \implies \dfrac{u_{n+1}}{u_n}=\dfrac{x^{n+1}n!}{x^n(n+1)!}=\dfrac{x}{n+1}$

$\therefore\displaystyle\lim_{n \to \infty}\dfrac{u_{n+1}}{u_n}=0$

$\therefore\displaystyle\lim_{n \to \infty}u_n=0$

Since for a sequence $\{u_n\}$ of positive real numbers such that $\displaystyle\lim_{n \to \infty}\dfrac{u_{n+1}}{u_n}=L\ (<1)$ we must have $\displaystyle\lim_{n \to \infty}u_n=0$.

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Lemma: Let $u_n>0$ and $v_n>0$ such that ; there exists $N$ st for all $n\geq N$; $\dfrac{u_{n+1}}{u_n}\leq \dfrac{v_{n+1}}{v_n}$. Then the sequence $\dfrac{u_n}{v_n}$ is bounded.

Proof: for all $n\geq N$ we have $\dfrac{u_{n+1}}{v_{n+1}}\leq \dfrac{u_n}{v_n}$ hence the sequence $(\dfrac{u_n}{v_n})_{n\geq N}$ is decreasing in particular it is bounded (it is positive). This show also that the sequence $(\dfrac{u_n}{v_n})$ is bounded.

Application: let $x\in \Bbb R^*$. Let $u_n=(2|x|)^n$ and $v_n=n!$. We have $\dfrac{u_{n+1}}{u_n}=2|x|$ and $\dfrac{v_{n+1}}{v_n}=n+1$. Now for $N=[2x]$ we have: $\forall n\geq N$ ; $\dfrac{u_{n+1}}{u_n}=2|x|\leq N+1\leq n+1=\dfrac{v_{n+1}}{v_n}$. It follos that the sequence $\dfrac{u_n}{v_n}$ is bounded, then there exists $M\in \Bbb R^+$ such that $\dfrac{u_n}{v_n}\leq M$ i.e $0\leq \dfrac{|x|^n}{n!}\leq \dfrac{M}{2^n}$ so $\lim_{n\to +\infty}\dfrac{|x|^n}{n!}=0$ thus $\lim_{n\to +\infty}\dfrac{|x|^n}{n!}=0$.

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