1

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.

Raghu
  • 195
  • 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 05 '11 at 07:14

1 Answers1

4

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 $n$?

Steven Stadnicki
  • 51,746
  • I don't know the $\epsilon - \delta$ definition of a 'limit'. I do know the $\epsilon$ definition. – Raghu Oct 05 '11 at 07:06
  • 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 05 '11 at 14:28