Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
I guess you mean $\lim_{n\to\infty}$ for fixed $x$? – Fabian Apr 7 '11 at 17:52
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 $0$.

share|cite|improve this answer

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.