Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\displaystyle\lim_{n\to\infty} \dfrac{1}{n!} = 0$

I have no idea how to proceed with this. Usually I start with a preliminary computation and solve for $n$ in terms of $\epsilon$ from the definition of the limit:

$|\dfrac{1}{n!} - 0| < \epsilon$

Here, I do not see an approach to solve for n.

share|improve this question
    
Solving for $n$ is totally unnecessary. You are not asked to find the smallest $N$ such that beyond it $|a_n|\lt \epsilon$. You are only asked to show there is an $N$ such that beyond it, $\dots$. –  André Nicolas Sep 29 '12 at 23:00
add comment

2 Answers

up vote 3 down vote accepted

HINT: $$\frac1{n!}\le\frac1n$$ for $n>0$. Now use the Archimedean property of the reals.

share|improve this answer
add comment

You can either try to find a $n$ for which $$ n!\geq \frac{1}{\epsilon} $$ or as Brian suggested, you only need to find $n$ such that $$ n\geq \frac{1}{\epsilon}. $$ In either case, it proves that the limit goes to $0$.

share|improve this answer
add comment

Your Answer

 
discard

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.