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Suppose $\lim\limits_{n \rightarrow \infty} a_n=A$. I proved $\lim\limits_{n \rightarrow \infty} a_n^p=A^p$ for $p \in N$ using induction. Is there an easier way for doing this? And how can I prove $\lim\limits_{n \rightarrow \infty} a_n^{1/p}=A^{1/p}$ for $p \in N$?

Regards,

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$x\mapsto x^p$ is always continuous. (Or is that what you're aiming at in the first place?) –  Henning Makholm Oct 27 '11 at 17:08
    
Did you use the $\epsilon-\delta$ definition in your induction? –  JavaMan Oct 27 '11 at 17:10

2 Answers 2

up vote 2 down vote accepted

You can't quite prove the latter, because the expressions may not make sense (e.g., if $a_n\lt 0$ and $p$ is even).

So let us assume that $a_n\geq 0$ if $p$ is even.

You need to show that for every $\epsilon\gt 0$ there is an $N\gt 0$ such that if $m\geq n$, then $|a_m^{1/p} - A^{1/p}|\lt\epsilon$.

Now, notice that $$(x-y)(x^r + x^{r-1}y + x^{r-2}y^2 + \cdots + xy^{r-1} + y^r) = x^{r+1}-y^{r+1}.$$ So $$a_m^{1/p} - A^{1/p} = \frac{a_m - A}{a_{m}^{(p-1)/p} + a_m^{(p-2)/p}A^{1/p} + \cdots + a_m^{1/p}A^{(p-2)/p} + A^{(p-1)/p}}.$$ We know we can make $a_m-A$ as small as we want provided we take $m$ large enough.

Also, if we take $m$ large enough, then $A-1 \leq a_{m} \leq A+1$ (or some other suitable constant so that $0\lt A-1$), so taking appropriate powers we can bound $a_m^{k}$ by an expression that depends only on $A$ and $k$. Putting all of these together, we can bound $$\left|\frac{1}{ a_{m}^{(p-1)/p} + a_m^{(p-2)/p}A^{1/p} + \cdots + a_m^{1/p}A^{(p-2)/p} + A^{(p-1)/p}}\right|$$ from above by an expression that depends only on $A$ and $p$. So it should be possible to find a sufficiently large $m$ so that this bound holds, and we can make $|a_m-A|$ so small that we know that $|a_m^{1/p}-A^{1/p}|$ is smaller than the desired $\epsilon$.

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Yes, finally found my own answer :-): is this legal?

$\lim_{n \to \infty} a_n^{1/p}$. Now choose $b_n^p=a_n$. Then $\lim_{n \to \infty} a_n^{1/p}=\lim_{n \to \infty}b_n$. Now $(\lim_{n \to \infty} b_n)^p=\lim_{n \to \infty}b_n^p=\lim_{n \to \infty} a_n=A$. So $(\lim_{n \to \infty} b_n)^p=A$. And so $\lim_{n \to \infty} a_n^{1/p}=\lim_{n \to \infty}b_n=A^{1/p}$ what was asked! :-)

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Yes, answering your own question is allowed. But, it seems you're only proving that if the limit exists, then it must be $A^{1/p}$. –  Henning Makholm Oct 27 '11 at 18:12

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