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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.2011-12-12
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    Which gives $(x-a) \frac{a_{k+1}}{a_k} \le (x-a) |\frac{a_{k+1}}{a_k}|$, but what to do next?2011-12-12
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    I gave an answer; but try to work it out from just the initial hint. :)2011-12-12

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