# How can I calculate the limit of exponential divided by factorial?

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

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

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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

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.

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

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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 $(2)$.

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

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

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