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I'm supposed to show that

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

without using any formulas for factoriel like Stirling's formula. I can see that by ploting the functions but I can't figure an algebric solution.

Alex. N
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  • $\dfrac{e^{x+1}}{(x+1)!}=\dfrac{e^x}{x!}\cdot\dfrac e{x+1}$, so you are eventually multiplying a fixed $\dfrac{e^{K}}{K!}$ by numbers which are $\le \dfrac 12$... –  Nov 03 '17 at 07:00

1 Answers1

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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.

Aditya
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