Given $a\in\mathbb{R}$ and $0
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.
Simple Limit Proof
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limits
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0Do you know of logarithms? – 2012-04-05
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0@anon No, we haven't gotten to logarithms yet. – 2012-04-05
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0What "limit definition" do you have? The one with $\epsilon$ ...? – 2012-04-05
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0@ThomasM $\forall \epsilon>0, \exists n_0$, $\forall n\ge n_0$, $|a^n|<\epsilon$ – 2012-04-05
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0@Anonymous In that case Andre's answer will work fine... – 2012-04-05
1 Answers
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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
$$0
Remark: One could use fancier tools. The sequence $(a^n)$ is decreasing. It is bounded below by $0$. So the sequence has a limit. Let $L$ be the limit. Then
$$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 $L=0$.
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0why does $lim_{n\to\infty} a^n=\lim_{n\to\infty}a^{n+1}$? – 2012-04-05
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0Anonymous also wishes to note that your first inequality is [Bernoulli's inequality](http://en.wikipedia.org/wiki/Bernoulli's_inequality). – 2012-04-05
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0Definition of limit. If $|a^n-L|<\epsilon$ whenever $n >N$, then $|a^{n+1}-L|<\epsilon$ for $n>\epsilon$. – 2012-04-05
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0I'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. – 2012-04-05
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0Sorry, 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 $n>N$. – 2012-04-05