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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 $g(z)=\sum_0^\infty a_n (z-z_0)^n$

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 had trouble writing the coefficients in the form $g(z)=\sum_0^\infty b_nz^n$.

Any suggestions? Note: Not a hw 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
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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. –  copper.hat Oct 11 '12 at 4:17

2 Answers 2

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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