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 $R-r$.
Best Answer
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, $$\eqalign{|b_k| s^k &\le \sum_{n=k}^\infty |a_n| {n \choose k} r^{n-k} s^k\cr &\le \sum_{n=k}^\infty B (s+r)^{-n} {n \choose k} r^{n-k} s^k \cr &= \sum_{j=0}^\infty B (s+r)^{-j-k} {j+k \choose k} r^j s^k\cr &= B \left(\frac{s}{s+r}\right)^k \left(1 - \frac{r}{s+r}\right)^{-k-1} = B \frac{s+r}{s}\cr}$$