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As the limit of $n$ goes to infinity, prove that $x^n = 0$ if $\operatorname{abs}(x)<1$.

So I want to prove it this by observing that $\operatorname{abs}(x) < 1$ which means $1/(\operatorname{abs}(x)) > 1$. How would I proceed from here?

Mike Pierce
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user2097612
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    This is usually taken as obvious, if you just observe what would happen. Are you just trying to find some way of showing it ? – Quality Apr 26 '15 at 19:12
  • A common strategy is show $\lim_{n\to\infty} |x|^n = 0$ by demonstrating that the sequence $|x|^n$ is decreasing and bounded below by $0$.

    It then follows that $\lim_{n\to\infty} x^n = 0$.

     – Simon S Apr 26 '15 at 19:17
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    Simon S has pointed out a way to see that it converges, not why it converges to $0$. – Jonas Meyer Apr 26 '15 at 19:18
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    http://math.stackexchange.com/questions/126042/proving-that-xn-converges-to-0-whenever-x-1

    Just complete Simon's argument or must you use that condition in the question?

     – Ilham Apr 26 '15 at 19:21
  • Yes, by using the condition 1/(abs(x)) > 1 – user2097612 Apr 26 '15 at 19:21
  • I must use that condition in the question. – user2097612 Apr 26 '15 at 19:27
  • Then just use Simon's argument to note that the sequence converges to some $c$. Suppose $c\neq 0$. Then note that $\lim_{n \to \infty} x^{n+1} = \lim_{n \to \infty} x^n = c$, so $cx= c$, and so $|c||x| = |c|$ and so $\frac{|c|}{|x|} = |c|$ but this cannot be since $\frac{1}{|x|} >1$. – Ilham Apr 26 '15 at 19:27

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So you want to focus on $1/|x|$ being larger than $1$. Let $a=1/|x|$. Because $a>1$, we have $a=1+h$ with $h>0$. There exists a positive integer $k$ such that $kh>1$. By Bernoulli's inequality, $a^k=(1+h)^k\geq 1+hk>2$. Then for each positive integer $m$, $a^{km}>2^m>m$. This implies that $\lim\limits_{m\to\infty}a^{km}=+\infty$, which implies that $\lim\limits_{n\to\infty}a^n=+\infty$ because the sequence is monotone increasing. This in turn implies that $\lim\limits_{n\to\infty}\dfrac{1}{a^n}=\lim\limits_{n\to\infty}|x|^n = 0$.

Jonas Meyer
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