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Why is $\lim\limits_{N\to\infty}x^{N+1}=0$, where $-1<x<1$?

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Have you tried applying the definition? – chubakueno Apr 24 '14 at 21:02
I don't get what it means by the |x| <1 – Manuel Apr 24 '14 at 21:06
$|x|<1$ is the same as saying $-1<x<1$. Sorry about the confusion. I'll change it back. – Sujaan Kunalan Apr 24 '14 at 21:08
@chubakueno Hola otra vez! como estas? – ProbabilityGuy Apr 24 '14 at 21:10
Oh okay...if $N$ keeps growing to Infinity. Shouldn't the limit be $\infty$? – Manuel Apr 24 '14 at 21:11
up vote 2 down vote accepted

Consider thinking about it intuitively.

We know that $-1<x<1$. Let's take a arbitrary $x$ as $x=\frac{1}{2}$.

Now $(\frac{1}{2})^1=\frac{1}{2}=0.500$,




As $n$ gets larger and larger, our value will get smaller and smaller. This is because $-1<x<1$, so raising it to $n^{th}$ power as $n$ gets larger and larger actually makes our value smaller and smaller.

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I see. I didn't take it into account that x was limited between -1 and 1. Thanks – Manuel Apr 24 '14 at 21:21

Here is a trick:

Put $|x| = \frac{1}{1 + y} < 1$. We want to show that $x^n \to 0 $. In other words, by definition, for a given $\epsilon > 0$, we want to find an $N$ such that for all $n \geq N$, then $|x^n - 0 | < \epsilon $. To find $N$, notice by Bernoulli's inequality

$$ |x^n| = \frac{1}{(1+y)^n} \leq \frac{1}{1+yn} <\epsilon \iff n \geq \frac{1 - \epsilon}{\epsilon y}$$

So, choosing $N = $ integer part of $\frac{1 - \epsilon}{y \epsilon} $ gives desired result!

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(+1) I didn't see your answer when I started writing mine. I see that it is essentially the same. Bernoulli is the way to go to avoid needing logarithms. I will remove my answer if you think it should be. – robjohn Apr 24 '14 at 21:45
No, I think your answer is written better than mine! – ProbabilityGuy Apr 24 '14 at 21:46

If $|x|\lt1$, choose an integer $k\ge\frac{|x|}{1-|x|}$. Then $$ |x|\le\frac{k}{k+1} $$ Bernoulli's Inequality says that $\left(1+\frac1k\right)^n\ge1+\frac nk$. Therefore, $$ |x|^n\le\left(\frac{k}{k+1}\right)^n\le\frac{k}{k+n} $$ If we wish to make $|x|^n$ smaller than any given $\epsilon\gt0$, choose $n\ge k/\epsilon$. Then, $$ |x|^n\le\frac{k}{k+n}\le\frac{k}{k+k/\epsilon}=\epsilon\frac{k}{k\epsilon+k}\le\epsilon $$

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A relatively "low-tech" way to see the limit must be zero (assuming the limit exists) is to call the limit $L$ and note that $$ L = \lim_{N \to \infty} x^{N+1} = \lim_{N \to \infty} (x \cdot x^{N}) = x \lim_{N \to \infty} x^{N} = xL. $$ Subtracting and factoring, $(1 - x)L = 0$. Since $x \neq 1$ by hypothesis, it must be that $L = 0$.

To prove the limit exists without descending into $\varepsilon$-land (i.e., using convergence criteria that appear early in an elementary analysis course), note that if $0 \leq x < 1$, then the sequence $(x^{N})_{N=0}^{\infty}$ is non-increasing and bounded below by $0$, so it has a real limit $L$.

To handle the case $-1 < x < 0$, note that $-|x|^{N} \leq x^{N} \leq |x|^{N}$ for all $N \geq 0$ and apply the squeeze theorem.

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Given $x \neq 0$ (case equal to zero is trivial) you want $|x^n|<\epsilon \ \ \forall N>n$ for some $N$ you have to find that $N$. $$|x^n|<\epsilon \iff|x|^n<\epsilon \iff n \log|x|<\log\epsilon \iff n>\frac{\log \epsilon}{\log|x|} $$

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the last implication is not true – ProbabilityGuy Apr 24 '14 at 21:43
@Lemur why si not true? – rlartiga Apr 24 '14 at 23:28

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