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 $L=\infty, R=0$.

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Hint: apply the ratio test. – David Mitra Dec 12 '11 at 9:02
Which gives $(x-a) \frac{a_{k+1}}{a_k} \le (x-a) |\frac{a_{k+1}}{a_k}|$, but what to do next? – Kevin Dec 12 '11 at 9:07
I gave an answer; but try to work it out from just the initial hint. :) – David Mitra Dec 12 '11 at 9:17

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 $1\over L$.

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