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Say, $a_n → \alpha$. ($\alpha \in \mathbb{R}$)

How do I prove that the sequence $\{a_n ^r\}$ converges to $\alpha ^r$ when $r\in \mathbb{R}$? (Only if $\alpha ^r$ can be defined)

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Is this true? What if $\alpha = 0$ and $r = -1$? –  Matt N. Sep 20 '12 at 8:42
    
How do you define $\alpha^r$ when $\alpha\in\mathbb C$ and $r\in\mathbb R$? –  Marc van Leeuwen Sep 20 '12 at 8:42
    
Edited! My mistake –  Katlus Sep 20 '12 at 8:44
    
This follows from the fact the $f(x)=x^r$ is continuous, (except at $x=0$ when $r<0$). –  Teddy Sep 20 '12 at 8:47
    
@Teddy Is there any direct proof for at least when $r\in \mathbb{Q}$? –  Katlus Sep 20 '12 at 8:50

1 Answer 1

up vote 0 down vote accepted

For $r\in\mathbb Q$, you should consider 3 cases:

  1. $r\in\mathbb N$: use $(\alpha_n)^r-\alpha^r = (\alpha_n - \alpha)\cdot\displaystyle\sum_{k=0}^{r-1}{\alpha_n}^k\alpha^{r-1-k}$
  2. $r=\displaystyle\frac 1b,\ b\in\mathbb N$: use the same backwards for $\sqrt[b]{\alpha_n}-\sqrt[b]{\alpha}$
  3. $r=-1$: use $\displaystyle\frac1{\alpha_n}-\frac1\alpha = \frac{\alpha-\alpha_n}{\alpha_n\alpha}$. Here $\alpha\ne 0$ (hence for almost all $n$, $|\alpha_n|$ is above a positive limit) is needed.

For arbitrary $r\in\mathbb R$, in a sense you have to use continuity, anyway exactly continuity defines $\alpha^r$ for all $r\in\mathbb R$..

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