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.

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.

share|improve this question
1  
What have you tried? Where are you getting stuck? Is this homework? If it is, please tag it as homework. –  Matthew Conroy Apr 5 '12 at 21:46
    
Hey, no it's not homework, it's something my professor assumed was obvious but he didn't prove it. –  Anonymous Apr 5 '12 at 21:46
    
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

1 Answer 1

up vote 2 down vote accepted

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 $L=0$.

share|improve this answer
    
why didn't you write $\frac{1}{(1+k)^n} \le \frac{1}{1+kn}$? Thanks a lot. –  Anonymous Apr 5 '12 at 22:08
    
@Anonymous: Thanks, fixed. –  André Nicolas Apr 5 '12 at 22:10
    
why does $lim_{n\to\infty} a^n=\lim_{n\to\infty}a^{n+1}$? –  Anonymous Apr 5 '12 at 22:14
    
Anonymous also wishes to note that your first inequality is Bernoulli's inequality. –  anon Apr 5 '12 at 22:21
    
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

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.