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I've already proved that for any $p > 0$ and for any $\alpha \in \mathbb{R}$, the sequence $\frac{n^\alpha}{(1 + p)^n}$ converges to $0$.

Now, I want to prove that $\lim_{n \to \infty} x^n = 0$ as long as $|x| < 1$.

I've split it into two possible cases:

(1) $0 < x < 1$. Let $p = \frac{1}{x} - 1$, so that $x = \frac{1}{1 + p}$ and $p > 0$ (since $x < 1$). Then we let $\alpha = 0$ to see that $$\lim_{n \to \infty} x^n = \lim_{n \to \infty} \frac{n^0}{(1 + p)^n} = 0$$

(2) $-1 < x < 0$. Here we cannot use the sequence $\frac{n^\alpha}{(1 + p)^n}$ again, since if we let $p = \frac{1}{x} - 1$, then $p$ isn't necessarily positive (for example, if $x = - \frac{1}{2}$, then $p = -3 < 0$).

How can I prove Case (2)?

jamaicanworm
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2 Answers2

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It suffices to show that $|x|^n=|x^n| \to 0$, so we can suppose $0\leq x <1$. Clearly the sequence is decreasing and bounded below by $0$, so it converges, say to $C$. Since $x$ is constant, then $C = \lim_{n\to\infty}{x^{n+1}} = x \cdot \lim_{n\to\infty}{x^n}=x \cdot C$, and since $x \neq 1$ we must have $C=0$.

Derek Pan
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Bruno Joyal
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    Do you mean bounded below by zero? – user193319 Sep 02 '18 at 15:09
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    Also, how did you deduce $C = xC$? The best I could come up with is the following: since $(x_n)$ converges to $C$, so must the subsequence $(x^{n+1})$. Therefore $C = \lim_{n \to \infty} x^{n+1} = x \lim_{n \to \infty} x^n = xC$. Is that a correct way of deducing it? – user193319 Sep 02 '18 at 15:16
  • @user193319 yes and yes. – Bruno Joyal Sep 06 '18 at 05:45
  • @user193319 or you can use the difinition of limit to show that xC is also a limit and conclude by uniqueness of limit – Marco Flores Apr 21 '21 at 10:38
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In your second case put $p = -\frac{1}{x} - 1$ , now p is always positive and conclude your result.

quartz
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  • This doesn't work. $p = - \frac{1}{x} - 1$ gives us $x = -\frac{1}{1 + p}$, so that $x^n = (-1)^n \frac{1}{1 + p}$, whose convergence we are not certain of. – jamaicanworm Mar 29 '12 at 20:11
  • as n tends to infinity value of $x^n = (-1)^n (\frac{1}{1 + p})^n$ tends to 0 anyway. We are not certain of convergence of $(-1)^n$, but we are always certain of convergence of $(-1)^n (\frac{1}{1 + p})^n$ – quartz Mar 29 '12 at 20:18
  • Yes, but this requires additional steps in the proof. – jamaicanworm Mar 30 '12 at 01:18