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

Question

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

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

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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.

Answer 1

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 $k\geq 7>3$.

Answer 2

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 $0$.

Answer 3

I have deleted my previous approach to the first question because it was substandard. Instead, for $n\ge2$, we have $$ \frac{2^n}{n!}=\frac{\overbrace{2\cdot2\cdot2\cdots2}^{\text{$n$ copies}}}{1\cdot2\cdot3\cdots n}\le\frac{2\cdot2}{1\cdot2}\left(\frac23\right)^{n-2}\to0\qquad\text{as }n\to\infty $$

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

Answer 4

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.

Answer 5

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 $\lim_n a_n =0$.

Answer 6

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.