Show that $\sum_{k=1}^{\infty} kx^{k-1}$ converges uniformly on $[a,b]$ for any $-1

I wish to show that $\sum_{k=1}^{\infty} kx^{k-1}$ converges uniformly on $[a,b]$ for any $-1<a<b<1$.

Clearly this series is the derivative of the geometric series , so my thought is to use the Weierstrass M test so that $kx^{k-1} < kb^{k-1}$. Is is then true that $\sum_{k=1}^{\infty}kb^{k-1}$ converges because it is the derivative of $\sum_{k=1}^{\infty}b^{k}$ which is convergent or perhaps I am on the wrong track. Hints appreciated.