2
$\begingroup$

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

  • 1
    I guess you mean $\lim_{n\to\infty}$ for fixed $x$?2011-04-07

2 Answers 2

5

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|. 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).