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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.
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 $k\geq 7>3$.
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 $0$.
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 $$ \lim_{n\to\infty}\frac{x^n}{n!}=0 $$
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 $\lim_n a_n =0$.
Define the sequence $\{ a_n\}$ as $a_n= \dfrac{x^n}{n!}$ for $x\in \mathbb R$ and $n\in \mathbb N$.
If $x=0$, it is trivial that $\lim a_n=0$
If $x>0$, then one has that
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.
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.
$u_n=\dfrac{x^n}{n!} \implies \dfrac{u_{n+1}}{u_n}=\dfrac{x^{n+1}n!}{x^n(n+1)!}=\dfrac{x}{n+1}$
$\therefore\displaystyle\lim_{n \to \infty}\dfrac{u_{n+1}}{u_n}=0$
$\therefore\displaystyle\lim_{n \to \infty}u_n=0$
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 $\displaystyle\lim_{n \to \infty}u_n=0$.
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.
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 $\displaystyle \lim_{n\to\infty}\frac{x^n}{n!}=0$
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 $$|x^n| =e^{n\ln |x|}$$
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.
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 $\lim_{n\to +\infty}\dfrac{|x|^n}{n!}=0$.
$\dfrac{2}{n!} \leq \dfrac{2^n}{n!} \leq \dfrac{2^n}{3^n}$
By Squeeze Theorem, conclusion follows.
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 $\displaystyle \lim_{n \to\infty}\dfrac{|a|^n}{n!}=0.$
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 $n \rightarrow \infty$
