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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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    $\begingroup$ 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... $\endgroup$ – J. M. isn't a mathematician Oct 31 '11 at 16:59
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    $\begingroup$ 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! $\endgroup$ – Matt Nashra Oct 31 '11 at 17:05
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    $\begingroup$ Then the sequence converges, but not necessarily to zero. $\endgroup$ – AMPerrine Oct 31 '11 at 17:06
  • $\begingroup$ 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 $\frac{2^n}{n!} \le \frac{8}{6}\frac{1}{2^{n-3}}$. $\endgroup$ – André Nicolas Oct 31 '11 at 17:07
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    $\begingroup$ @JM: $0<\frac{1}{2}+2^{-(k+1)}<\frac{1}{2}+2^{-k}$, but that sequence does not converge to $0$. $\endgroup$ – robjohn Oct 31 '11 at 17:39

15 Answers 15

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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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    $\begingroup$ you should prove that for $n=7$ it works... :P $\endgroup$ – Valerio Capraro Oct 31 '11 at 17:55
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    $\begingroup$ @Valerio At this stage, it is conventional to say "It can be trivially verified that the inequality holds for $n=7$." :-) $\endgroup$ – Srivatsan Oct 31 '11 at 18:18
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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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I have deleted my previous approach to the first question because it was substandard. Instead, for $n\ge2$, we have $$ \frac{2^n}{n!}=\frac{\overbrace{2\cdot2\cdot2\cdots2}^{\text{$n$ copies}}}{1\cdot2\cdot3\cdots n}\le\frac{2\cdot2}{1\cdot2}\left(\frac23\right)^{n-2}\to0\qquad\text{as }n\to\infty $$


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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  • $\begingroup$ 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 $\endgroup$ – Ilya Nov 28 '14 at 12:06
  • $\begingroup$ @Ilya: I've moved my answer from the duplicate question here since it answers the second part. $\endgroup$ – robjohn Nov 28 '14 at 14:09
  • $\begingroup$ Now I can happily upvote it :) $\endgroup$ – Ilya Nov 28 '14 at 14:11
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I am surprised that no one 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 we have

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

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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  • $\begingroup$ This one is my favorite because it is a very simple and elegant explanation! $\endgroup$ – Argon Apr 20 '12 at 1:44
  • $\begingroup$ @N.S. That is a very nice idea. Would you mind generalizing it to any exponent $x$? $\endgroup$ – Pedro Tamaroff Apr 20 '12 at 1:45
  • $\begingroup$ @PeterT.off Done $\endgroup$ – N. S. Apr 20 '12 at 1:56
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    $\begingroup$ @N.S. Great! Now we have two proofs. Maybe someone else can complete the "Gaussian-like" triad =D. $\endgroup$ – Pedro Tamaroff Apr 20 '12 at 1:58
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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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    $\begingroup$ Any feedback on the downvote? $\endgroup$ – Pedro Tamaroff Apr 19 '12 at 19:04
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    $\begingroup$ 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. $\endgroup$ – Kaz Apr 19 '12 at 23:30
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    $\begingroup$ 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. $\endgroup$ – anon Apr 19 '12 at 23:31
  • $\begingroup$ @anon I'm polishing it. I generalized something which wasn't correct. $\endgroup$ – Pedro Tamaroff Apr 19 '12 at 23:32
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    $\begingroup$ 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? $\endgroup$ – Aryabhata Apr 19 '12 at 23:41
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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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  • $\begingroup$ Can you provide me the link of the proof of that theorem ? $\endgroup$ – NewBornMATH Feb 23 '19 at 13:43
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    $\begingroup$ See here. $\endgroup$ – user170039 Feb 23 '19 at 14:08
  • $\begingroup$ I think that(ratio test) is for series , which says if $L<1$ then the series converge,but my point is how you can guarantee that $\lim_{n->\infty} u_{n}=0$ from $L<1$. What i am thinking is there is a theorem i dont know the exact name but some people tell it cauchy's theorem on limit. Which says $\lim_{n->\infty} \frac{u_{n+1}}{u_{n}}=\lim_{n->\infty} (u_{n})^{1/n}$ is there any way to prove from this theorem ? $\endgroup$ – NewBornMATH Feb 23 '19 at 15:25
  • $\begingroup$ @NewBornMATH: If the series $\displaystyle\sum_{i=0}^\infty u_i$ converges then by Cauchy's theorem it follows that $\lim_{n\to\infty} u_n=0$. This answers your first question (unless you are asking for a proof of the theorem for the case $L<1$ - in which case I ask you to read the Wikipedia article again). For the second I would say that I don't know the answer to it. $\endgroup$ – user170039 Feb 23 '19 at 15:48
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    $\begingroup$ Sad this isn't upvoted more! $\endgroup$ – Simply Beautiful Art Dec 15 '19 at 19:02
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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: 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 $$\color{maroon}{0}\color{red}{<}\color{fuchsia}{P_n}\color{red}{<}\color{maroon}{\frac x1.\frac x2\cdots\frac{x}{x-1}.\frac xx.}\color{green}{\frac x{x+1}.\frac x{x+1}\cdots\frac{x}{x+1}.\frac x{x+1}}$$ 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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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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  • $\begingroup$ very nice proof +1,I do like it $\endgroup$ – Khosrotash Sep 25 '19 at 19:00
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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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First Answer The series $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$ converges then $$\lim_{n\to\infty}\frac{x^n}{n!}= 0$$

OR Second Answer Use the following famous Stirling formula: Given $x>0$ $$ \lim_{n\to +\infty} \frac{n!}{\left(\frac{n}{e}\right)^n\sqrt{2n} }=\sqrt{\pi}. $$ and $$|x^n| =e^{n\ln |x|}$$

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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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$\dfrac{2}{n!} \leq \dfrac{2^n}{n!} \leq \dfrac{2^n}{3^n}$

By Squeeze Theorem, conclusion follows.

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  • $\begingroup$ When $n\ge 7$, $n\in\mathbb Z$. $\endgroup$ – user236182 Sep 12 '17 at 22:06
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Since $\forall \, n \in \mathbb{N}$ we have $\bigg(1+\dfrac{1}{n}\bigg)^{n}<e,$ by induction it follows that $\bigg(\dfrac{n}{e}\bigg)^n<n!.$ For a fixed $a \in \mathbb{R}, $ by the Archimedean property, there exists $k \in \mathbb{N}$ such that $e(|a|+1)<k.$ Then for all $n \geq k, $ and by the second inequality we have we have $(|a|+1)<\dfrac{n}{e}<(n!)^{1/n}$ from which it follows that $0< \dfrac{|a|^n}{n!}\leq \dfrac{|a|^n}{(1+|a|)^n}$ which by the squeeze theorem forces $\displaystyle \lim_{n \to\infty}\dfrac{|a|^n}{n!}=0.$

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Since $\frac{-|x|^n}{n!} \leq \frac{x^n}{n!} \leq \frac{|x|^n}{n!}$

Therefore, enough to show that $\frac{|x|^{n}}{n!} \rightarrow 0$ as $n \rightarrow \infty$

Now $x$ is any real number, we can find $M$ such that $|x| < M$

which means $\frac{|x|}{M} < 1$, $\big(\frac{|x|}{M}\big)^n \rightarrow 0$

Now for all $n > M$ we have $\frac{|x|^n}{n!} = \frac{|x|^n}{1.2.3....M(M+1)...n} $ $\leq \frac{|x|^n}{M!M^{n-M}} $ = $\big(\frac{|x|}{M!}\big)^n \frac{M^M}{M^n} $

which will tend to $0$ as $n \rightarrow \infty$

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