The only thing you have to do is apply the Ratio test to the terms ${|a_n(x-a)^n|}$. Everything will drop out fairly easily by using the definition of the radius of convergence.
Warning: Solution follows:
Compute, for $x\ne a$: $\lim_{n\rightarrow\infty}{|a_{n+1}| |x-a|^{n+1}\over |a_n||x-a|^n } =\lim_{n\rightarrow\infty}{|a_{n+1}||x-a| \over |a_n| } =|x-a| \lim_{n\rightarrow\infty}{|a_{n+1}| \over |a_n| }=|x-a|L. $
The Ratio test allows you to conclude that the series converges whenever $|x-a|L<1$ and diverges whenever $|x-a|L>1$.
From the above, we can say:
If $L=0$, then the series converges for all $x$ and the radius of convergence is infinite.
If $L$ is infinite, then the series converges for no $x\ne a$. But the series does converge for $x=a$ (as trivially seen) and the radius of convergence is 0.
Otherwise, series converges whenever $|x-a| <{1\over L}$ and diverges whenever $|x-a| >{1\over L}$; which implies that the radius of convergence is $1\over L$.