Sign up ×
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.

This question already has an answer here:

I suspect this limit is 0, but how can I prove it?

$$\lim_{n \to +\infty} \frac{2^{n}}{n!}$$

share|cite|improve this question

marked as duplicate by Martin Sleziak, Macavity, Sami Ben Romdhane, Xoff, Cameron Buie Jan 15 '14 at 15:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Easy. If $n \geqslant 3$ then $\frac{2^n}{n!} \leqslant 2 \cdot \left(\frac{2}{3}\right)^{n-2}$. – Dan Shved Nov 20 '12 at 10:16

6 Answers 6

up vote 7 down vote accepted

The easiest way to do this is to do the following: Assume $n \ge 4$. Then $$0 \le \frac{2^n}{n!} = \prod_{i=1}^n \frac{2}{i} = \frac{2\cdot 2\cdot 2}{1 \cdot 2 \cdot 3} \cdot \prod_{i=4}^n \frac{2}{i} \le \frac{8}{6} \cdot \prod_{i=1}^n \frac{2}{4} = \frac{8}{6 \cdot 2^{n-3}}.$$ Applying the squeeze theorem gives the result.

share|cite|improve this answer

It's very easy to show by the ratio test that $\sum_{i=0}^\infty \frac{2^n}{n!}$ converges. It follows that the limit $\lim_{n\rightarrow\infty} \frac{2^n}{n!} = 0$.

share|cite|improve this answer

Suppose $n\ge4$ then $$n!=1\cdot 2\cdot 3 \cdot \underbrace{4}_{2\cdot 2}\cdot 5\cdot \cdots \cdot n\tag{1}$$ and $$2^n=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2 \cdots 2\tag{2}.$$

So every factor in $(1)$ is greater or equal than every factor in $(2)$.

share|cite|improve this answer

Hint: Show that $\displaystyle\frac{2^n}{n!} \le C\cdot\left(\frac 24\right)^n$ for almost all $n$.

share|cite|improve this answer

Using Stirling approximation $$ \lim_{n \to +\infty} \frac{2^{n}}{n!} = \lim_{n \to +\infty} \exp(\ln (2^n/n!)) = \lim_{n \to +\infty} \exp(n \ln 2 - n\ln n + n - O(\ln n )) = 0$$

share|cite|improve this answer

To prove $\lim_{n\to \infty} \,\frac{n!}{a^n} = +\infty$ it is suffice to prove that for all $k>0$ it is possible $\frac{n!}{a^n} > K$ for almost $n$, that is $n!>K a^n$. This means

$1\cdot 2\cdot\ldots\cdot\frac{n}{2}\cdot\ldots \cdot n > K \cdot a\cdot \stackrel{\stackrel{n}{\smile}}{\ldots}\cdot a,$

that is

$\frac{n/2+1}{a}\cdot\ldots\cdot \frac{n}{a} > K\cdot \frac{a}{1}\cdot\frac{a}{2}\cdot \stackrel{\stackrel{n/2}{\smile}}{\ldots}\cdot \frac{a}{n/2}. $

Note that

$\frac{n/2+1}{a}\cdot\ldots\cdot \frac{n}{a} > \left(\frac{n/2+1}{a}\right)^{n/2} > K \cdot a^{n/2} > K\cdot \frac{a}{1}\cdot\frac{a}{2}\cdot \stackrel{\stackrel{n/2}{\smile}}{\ldots}\cdot \frac{a}{n/2}$

which is true for $n$ verifying $\frac{n/2+1}{a} > 2 a,$ because $K^{2/n}=\sqrt[n]{K^2}\to 1$ and $2a>K^{2/n}a$ for $n$ large enough.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.