Prove that if $A_k \to A\,\, \left(A_k,\,\,A >0\right)$ then $\sqrt[n]{A_k} \to \sqrt[n]{A}$
I'm trying to prove the following result.
If
$$ \lim_{k\to\infty}A_k = A\,\,\,\,\,\, \left(A_k,\,\,A >0\right) $$
then
$$\lim_{k\to\infty}\sqrt[n]{A_k} = \sqrt[n]{A}$$
I am interested in using the hints given in the book I am reading, but I don't know how to continue, so please help me along the following lines. Use the identity
$$ x^n -x_0 ^n = \left(x-x_0\right)\left(x^{n-1}+x^{n-2}x_0+\ldots+x_0^{n-1}\right)$$
with $x=\sqrt[n]{A_k}$ and $x_0 = \sqrt[n]{A}$.
By replacing $x$ and $x_0$ we have $$ \left(\sqrt[n]{A_k}\right)^n - \left(\sqrt[n]{A}\right)^n = A_{k}-A$$ so $$ A_{k}-A= \left(\sqrt[n]{A_k} - \sqrt[n]{A}\right) \left(\left(\sqrt[n]{A_k}\right)^{n-1}+ \left(\sqrt[n]{A_k}\right)^{n-2}\left(\sqrt[n]{A}\right)+ \ldots+ \left(\sqrt[n]{A}\right)^{n-1} \right) $$ Then notice that all the terms in the sum are positive.....How to proceed?
Solution 1:
I think I have come up with a solution.
We have
$$ \vert\sqrt[n]{A_k} - \sqrt[n]{A} \vert=
\frac{
\vert A_k-A \vert}{\left(\sqrt[n]{A_k}\right)^{n-1}+
\left(\sqrt[n]{A_k}\right)^{n-2}\sqrt[n]{A}+
\ldots+
\left(\sqrt[n]{A}\right)^{n-1}
}
<
\frac{\vert A_k-A \vert}{\left(\sqrt[n]{A}\right)^{n-1}}$$
Then
$\vert\sqrt[n]{A_k} - \sqrt[n]{A} \vert \to 0 $
given that $ \vert A_k-A \vert \to 0$