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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 $n > x$?2011-04-07
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    I guess you mean $\lim_{n\to\infty}$ for fixed $x$?2011-04-07

2 Answers 2

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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|