What I want to show:
Let $\sum^{\infty}_{n=0} a_nx^n$ be a power series. Then $\sum^{\infty}_{n=0}na_nx^{n-1}$ has the same radius of convergence.
Proof:
Suppose $\sum^{\infty}_{n=0} a_nx^n$ converges for each $x_0\in [-r,r]$ where $0<r<R$ , for some $R>0$. That means $lim_{n\rightarrow \infty}$ $a_nx_0^n$ $=$ $0$, and so $\exists M_0$ such that $|a_nx_0^n|\leq M_0$ for each $n\in \mathbb{N}$. So if $x\in [-a,a]$, where $0<a<|x_0|$, we would have
$|n||a_n||x|^{n-1}$ $=$ $\frac{n}{a} |a_n||x_0|^n \frac{|a|^n}{|x_0|}$ $\leq$ $M_0\frac{n}{a}|\frac{a}{x_0}|^n$. Since each term in the following series:
$\sum^{\infty}_{n=1} M_0\frac{n}{a}|\frac{a}{x_0}|^n$
is non-zero, we may use the ratio test. Further, as $|a|<|x_0|$, the ratio test gives us a limit $L<1$ and so the series converges. Therefore, by the Weirstrass $M$ test, $\sum^{\infty}_{n=1}na_nx^{n-1}$ converges uniformly on $[-a,a]$.
But $[-a,a]\neq [-r,r]$, since $a<r$, so how do I complete the proof?
Here's a way out of your conundrum: Let $R>0$ be the radius of convergence of $\sum a_nx^n.$ Let $0<r<R.$ We want to show $\sum n|a_n|r^{n-1}<\infty.$
Choose any $s$ such that $r<s<R.$ Now use your approach, this time noting $\sum |a_n|s^n<\infty.$