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Given the sequence $\displaystyle\left\{\frac{x^n}{n!}\right\}$, how would I prove that its limit as $n\to\infty$ is zero?

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Hint: How is each term related to the one before it? What does this mean when $n > x$? –  Nate Eldredge Apr 7 '11 at 17:52
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I guess you mean $\lim_{n\to\infty}$ for fixed $x$? –  Fabian Apr 7 '11 at 17:52

2 Answers 2

up vote 5 down vote accepted

Consider the ratio of (absolute values of) the $n+1$st term by the $n$th term: $$\lim_{n\to\infty}\frac{\quad\frac{|x|^{n+1}}{(n+1)!}\quad}{\frac{|x|^n}{n!}} = \lim_{n\to\infty}\frac{|x|^{n+1}n!}{|x|^n(n+1)!} = \lim_{n\to\infty}\frac{|x|}{n+1} = 0.$$

Since the limit of the ratios is $0$, that means that the terms go to $0$.

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Choose $k$ large enough such that $|x|<k$. Then $$\frac{|x|^n}{n!} = \frac{|x|^k}{k!} \frac{|x|^{n-k}}{(k+1)(k+2)\dots n} \le \frac{|x|^k}{k!} \left(\frac{|x|}{k}\right)^{n-k}$$ The last term converges to 0 (geometric progression).

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