# Simple Limit Proof

Given $a\in\mathbb{R}$ and $0<a<1$, let $X_n=a^n$, $\forall n\in\mathbb{N}$.

Prove that $\lim \limits_{n\to \infty}X_n=0$ using limit definition or limits arithmetics(including the squeeze theorem if needed).

Thanks a lot.

-
Do you know of logarithms? – anon Apr 5 '12 at 21:50
@anon No, we haven't gotten to logarithms yet. – Anonymous Apr 5 '12 at 21:52
What "limit definition" do you have? The one with $\epsilon$ ...? – Thomas Apr 5 '12 at 21:54
@ThomasM $\forall \epsilon>0, \exists n_0$, $\forall n\ge n_0$, $|a^n|<\epsilon$ – Anonymous Apr 5 '12 at 22:00
@Anonymous In that case Andre's answer will work fine... – Thomas Apr 5 '12 at 22:03

Note that $\dfrac{1}{a}>1$. Let $\dfrac{1}{a}=1+k$.
By induction, or by using the Binomial Theorem, we can show that $(1+k)^n \ge 1+nk$. It follows that $$0<a^n=\frac{1}{(1+k)^n}\le\frac{1}{1+kn}.$$ Now it should not be hard to use the $\epsilon$-$N$ definition, or Squeezing, to get the result.
Remark: One could use fancier tools. The sequence $(a^n)$ is decreasing. It is bounded below by $0$. So the sequence has a limit. Let $L$ be the limit. Then $$L=\lim_{n\to\infty} a^n=\lim_{n\to\infty}a^{n+1}=a\lim_{n\to\infty} a^n=aL.$$ so $L(1-a)=0$ and therefore $L=0$.
why does $lim_{n\to\infty} a^n=\lim_{n\to\infty}a^{n+1}$? – Anonymous Apr 5 '12 at 22:14
Definition of limit. If $|a^n-L|<\epsilon$ whenever $n >N$, then $|a^{n+1}-L|<\epsilon$ for $n>\epsilon$. – André Nicolas Apr 5 '12 at 22:25
I'm sorry, but I didn't fully understand your last answer. Could you please explain again why from the definition of limit $|a^{n+1}-L|<\epsilon$ for $n>\epsilon$? Thanks a lot. – Anonymous Apr 5 '12 at 22:35
Sorry, typo, should be for $n>N$. Very informally, if we know that $|a^n-L|<0.001$ if $n>N$, it follows immediately that $|a^{n+1}-L|<0.001$ if $n>N$. – André Nicolas Apr 5 '12 at 22:40