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Suppose that $\sum_0^\infty a_nz^n$ has radius of convergence $1$ and suppose that $|z_0|=r<R$. 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 $g(z)$. – Robert Israel Oct 11 '12 at 4:17
The radius of convergence relates to the $\{a_n\}$. So if the first series has ROC $R$, then so will the shifted series. – copper.hat Oct 11 '12 at 4:17

The complex-analysis proof is easiest: $g(z) = \sum_n a_n (z - z_0)^n$ is analytic in the disk $\{z: |z - z_0|<R\}$, and therefore in the disk $\{z: |z|<R - r\}$ which it contains, so the radius of convergence is at least the radius $R - r$ of that disk.

But if you insist on a "real-analysis" proof: $$b_k = g^{(k)}(0)/k! = \sum_{n=k}^\infty a_n {n \choose k} (-z_0)^{n-k}$$ For $0 < s < R - r$, since $s+r < R$ we have $|a_n| (s+r)^n \to 0$ as $n \to \infty$. Take $B$ so $|a_n| (s+r)^n \le B$ for all $n$. Then using the binomial series, $$\eqalign{|b_k| s^k &\le \sum_{n=k}^\infty |a_n| {n \choose k} r^{n-k} s^k\cr &\le \sum_{n=k}^\infty B (s+r)^{-n} {n \choose k} r^{n-k} s^k \cr &= \sum_{j=0}^\infty B (s+r)^{-j-k} {j+k \choose k} r^j s^k\cr &= B \left(\frac{s}{s+r}\right)^k \left(1 - \frac{r}{s+r}\right)^{-k-1} = B \frac{s+r}{s}\cr}$$

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Wow, you type faster than I think... – copper.hat Oct 11 '12 at 4:19
I was writing up my answer and saw you post the complex-analytic proof. However, since the question was tagged real-analysis, I continued. I didn't see your "real" proof until after I finished mine. – robjohn Oct 11 '12 at 6:03
well, someone deserves a point for this answer. – robjohn Oct 16 '12 at 23:05
Thank you for posting this! It was helpful! – Froozle Oct 26 '12 at 1:54
@Froozle: If you find an answer helpful, consider upvoting it. – robjohn Oct 26 '12 at 5:37

Suppose the radius of convergence of $$ \sum_{n=0}^\infty a_nz^n\tag{1} $$ is $R$. This means that for any $r\lt R$, there is a $c_r$ so that $|a_n|\le c_r/r^n$.

Suppose we also have the series expanded about the point $z_0$: $$ \sum_{k=0}^\infty b_n(z-z_0)^n=\sum_{n=0}^\infty a_nz^n\tag{2} $$ Substituting $z\mapsto z+z_0$ yields $$ \begin{align} \sum_{k=0}^\infty b_nz^n &=\sum_{n=0}^\infty a_n(z+z_0)^n\\ &=\sum_{n=0}^\infty\sum_{k=0}^na_n\binom{n}{k}z^kz_0^{n-k}\\ &=\sum_{k=0}^\infty z^k\sum_{n=k}^\infty a_n\binom{n}{k}z_0^{n-k}\\ &=\sum_{k=0}^\infty z^k\sum_{n=0}^\infty a_{n+k}\binom{n+k}{k}z_0^n\tag{3} \end{align} $$ Thus, we get that for any $r\lt R$ (hence $r-|z_0|\lt R-|z_0|$), $$ \begin{align} |b_k| &=\left|\sum_{n=0}^\infty a_{n+k}\binom{n+k}{k}z_0^n\right|\\ &\le\sum_{n=0}^\infty\frac{c_r}{r^{k+n}}\binom{n+k}{n}|z_0|^n\\ &=\sum_{n=0}^\infty\frac{c_r}{r^{k+n}}(-1)^n\binom{-k-1}{n}|z_0|^n\\ &=\frac{c_r}{r^k}\left(1-\frac{|z_0|}{r}\right)^{-k-1}\\ &=\frac{c_rr}{r-z_0}\frac{1}{(r-|z_0|)^k}\tag{4} \end{align} $$ and $(4)$ says that $$ \sum_{n=0}^\infty b_nz^n\tag{5} $$ has a radius of convergence of at least $R-|z_0|$.

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Thank you! I started my proof this way but couldn't finish it...thanks for your help! – Froozle Oct 26 '12 at 1:54

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