3
$\begingroup$

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)?

  • 3
    Show that $|x|^n\to 0$, then use the fact that $-|x|^n\leq x^n \leq |x|^n$.2012-03-29

2 Answers 2

8

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 $C$. Then $C=xC$, and since $x \neq 1$ we must have $C=0$.

  • 0
    @user193319 yes and yes.2018-09-06
1

In your second case put $p = -\frac{1}{x} - 1$ , now p is always positive and conclude your result.

  • 0
    Yes, but this requires $a$dditional steps in the proof.2012-03-30