Proof about radius of convergence I have to proof this, but I don't know how to do this.
Let $R$ be the radius of convergence for $\sum_{k=0}^\infty a_k(x-a)^k$ and suppose that $\displaystyle\lim_{n \to +\infty} \left|\frac{a_{n+1}}{a_n}\right|=L$. Then:
(a) if $L$ is a nonzero finite real number, $R = \frac{1}{L}$, 
(b) if $L=0, R = \infty$,
(c) if $L=\infty, R=0$.

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 $1\over L$.
