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What is the epsilon proof that $x^n \rightarrow 0$ as $n \rightarrow \infty$ provided $|x| < 1 $? I only know it's true because I know the geometric series converges, which implies its terms must tend to 0, but never seen an epsilon proof of this simple fact.

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Let $|x|=\dfrac{1}{1+t}$. Since $|x|\lt 1$, we have $t\gt 0$.

By the Binomial Theorem, $(1+t)^n \ge 1+nt\gt nt$. Now it is easy to find $N$ such that if $n \gt N$, then $\dfrac{1}{nt}\lt \epsilon$.

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  • $\begingroup$ This may be a dumb question, but is there a broad theorem that says for any $x \in \mathbb{R}$, we can assume $x$ takes any form (in particular $x=\frac{1}{1+t}$ for $t \in \mathbb{R}, t \geq 0$)? $\endgroup$ – alwaysiamcaesar Apr 30 at 6:02
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Let us assume the limit of the sequence $\{x_n\}$ is $L$. Here $x_n=x^n$. We can easily show that limit exists by using Ratio Test (given $|x|<1$)

Since $L$ is the limit, we can always find an $N \in \mathbb{N}$ for every $\epsilon > 0\,$ s.t.
$$|x_n-L|<\epsilon \qquad \forall n\ge N$$

$\Rightarrow$$L-\epsilon <x^n<L+\epsilon$ and $L-\epsilon <x^{n+1}<L+\epsilon$

Also the above equaion imlies $(L-\epsilon)x <x^{n+1}<(L+\epsilon)x$

$\Rightarrow$ $L-\epsilon < (L+\epsilon)x$

$\Rightarrow$ $L<\epsilon \frac{1+x}{1-x}$

Now since $\epsilon$ can be arbitrarily small, you can easily show that $L=0$

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You don't need the binomial theorem. $(1+t)^n \ge 1+nt$ is Bernoulli's inequality, which is easily proved by induction on $n$. This proof is suitable for an introductory algebra class, once induction has been presented.

I first saw this proof in "What is Mathematics?" by Courant and Robbins - a wonderful book for learning math.

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    $\begingroup$ Shouldn't that rather be a comment to André's answer? $\endgroup$ – t.b. Sep 16 '12 at 4:01
  • $\begingroup$ Where's the answer? $\endgroup$ – Ravi Upadhyay Sep 16 '12 at 7:15

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