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.

Prove that if $\{a_n\}$ converges to $a$ and $|a| < 1$, then $\{(a_n)^n\}$ converges to 0.

This is what I have currently done. Please let me know if there is something wrong or if there is any other advice that you could provide to help me finish this.

Suppose $\{a_n\}$ converges to $a$ and $|a| < 1$. Since $\{a_n\}$ converges, then we know that $\{a_n\}$ is bounded. So there exists $M$ such that $|a_n| \leq M$. Then $-M \leq a_n \leq M$ for all $n$.

So I guess the problem that I am having is how to get M to be equal to 1. Thanks in advance.

share|improve this question
1  
For clarity let's take $a$ positive. Let $1-a=c$. Use the "$\epsilon$-$N$" definition of limit to show that there is an $N$ such that if $n >N$ then $|a_n-a|<c/2$. This forces $a_n$ to be less than $1-c/2$ (and, to be technical, $>-1/2$). Now large powers get us close to $0$. –  André Nicolas Oct 5 '11 at 7:14

1 Answer 1

up vote 4 down vote accepted

A hint: you haven't yet used the fact that the limit of the $a_n$, $|a|\lt 1$. Can you use that fact to be more specific about your bounds on $a_n$ for large $n$? A bigger hint: consider the classic definition of 'limit'; if you use an $\epsilon$ of ${1\over 2}(1-|a|)$, what can you say about the $a_n$ for (sufficiently) large $n$?

share|improve this answer
    
I don't know the $\epsilon - \delta$ definition of a 'limit'. I do know the $\epsilon$ definition. –  Raghu Oct 5 '11 at 7:06
    
Mea culpa - that was just a late-night 'thinko' on my part. As André says, that should be the '$\epsilon$-$N$' definition. –  Steven Stadnicki Oct 5 '11 at 14:28

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.