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Suppose that $\sum_0^\infty a_nz^n$ has radius of convergence $1$ and suppose that $|z_0|=r. Let $g(z)=\sum_0^\infty a_n (z-z_0)^n$.

Problem: Prove that $g(z)$ has radius of convergence at least $R-r$.

I saw this question in Beals and couldn't figure it out! I started expanding binomially, but I had trouble writing the coefficients in the form $g(z)=\sum_0^\infty b_nz^n$.

Any suggestions? Note: Not a homework problem. I am studying for a test on series and this was recommended for studying.

Edit: Hmm...maybe there is a typo. Though I'm not quite sure how an offcenter power series would still have radius of convergence $R$. To me, it seems somewhat intuitive that the radius of $g(z)$ would still be $R-r$.

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    Your statement of the question is not quite right. The series $\sum_n a_n (z - z_0)^n$ has the same radius of convergence as the series $\sum_n a_n z^n$. You're talking about the radius of convergence of the Maclaurin series $\sum_n b_n z^n$ of $g(z)$.2012-10-11
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    The radius of convergence relates to the $\{a_n\}$. So if the first series has ROC $R$, then so will the shifted series.2012-10-11

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