Given the sequence $\displaystyle\left\{\frac{x^n}{n!}\right\}$, how would I prove that its limit as $n\to\infty$ is zero?
How do I prove that this sequence has a limit of zero?
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$\begingroup$
calculus
sequences-and-series
limits
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1I guess you mean $\lim_{n\to\infty}$ for fixed $x$? – 2011-04-07
2 Answers
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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 $0$.
3
Choose $k$ large enough such that $|x|