Show that the series $\sum_{n=1}^{\infty}nx^n$ converges uniformly on [-a,a]

Question

So I'm trying to figure out how to show that $\sum_{n=1}^{\infty}nx^n$ converges uniformly on [-a,a], when a<1, but not on (-1,1) I've seen the same done for the regular geometric series $\sum_{n=0}^{\infty}x^n$, but I can't quite understand how to do it for this one. I know that by differentiating I can get the limit function for the series $$\frac{d}{dx}\sum_{n=0}^{\infty}x^n = \frac{d}{dx}\frac{1}{1-x}$$ $$\sum_{n=0}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2}$$ Multiply by $x$, $$\sum_{n=1}^{\infty}nx^n = \frac{x}{(1-x)^2}.$$ But how does the proof go?

Answer 1

Just use M-Test: \begin{align*} n|x|^{n}\leq n|a|^{n}, \end{align*} and we have $\displaystyle\sum n|a|^{n}<\infty$ for $|a|<1$ by Ratio Test.