Zorich's Mathematical Analysis Page 286:
- a) Making the formal substitution $z-a=(z-z_0)+(z_0-a)$ in the power series $\sum_{n=0}^{\infty}A_n(z-a)^n$ and gathering like terms, obtain a series $\sum_{n=0}^\infty C_n(z-z_0)^n$ and expressions for its oefficients in terms of $A_k$ and $(z_0-a)^k,k=0,1,2,\cdots$
b) Verify that if the original series converges in the disk $|z-a|<R$ and $|z_0-a|=r<R$ then the series defining $C_n,n=0,1,\cdots,$ converges absolutely and the series $\sum_{n=0}^\infty C_n(z-z_0)^n$ converges to $|z-z_0|<R-r$.
c) Show that if $f(z)=\sum_{n=0}^\infty A_n(z-a)^n$ in the disk $|z-a|<R$ and $|z_0-a|<R$, then in the disk $|z-z_0|<R-|z_0-a|$ the function $f$ admits the representation $f(z)=\sum_{n=0}^\infty C_n(z-z_0)^n$
I have managed to solve part $a)$, but have no idea for part $b)$.
My Attempt for part a)
$$I=\sum_{n=0}^{\infty}A_n(z-a)^n$$$$=\sum_{n=0}^\infty A_n[(z-z_0)+(z_0-a)]^n$$$$=\sum_{n=0}^\infty A_n\sum_{k=0}^n \binom{n}{k} (z_0-a)^{n-k}(z-z_0)^k$$$$=\sum_{n=0}^\infty A_n\sum_{k=0}^\infty \binom{n}{k} (z_0-a)^{n-k}(z-z_0)^k$$$$=\sum_{n=0}^\infty \sum_{k=0}^\infty A_n \binom {n}{k} (z_0-a)^{n-k}(z-z_0)^k$$$$\sum_{k=0}^\infty \sum_{n=k}^\infty A_n\binom {n}{k} (z_0-a)^{n-k}(z-z_0)^k$$$$\sum_{k=0}^\infty \left( \sum_{n=k}^\infty A_n\binom {n}{k} (z_0-a)^{n-k}\right)(z-z_0)^k$$
Which gives us $C_k= \sum_{n=k}^\infty A_n\binom {n}{k} (z_0-a)^{n-k}$
My Partial Attempt for part b)
We know by Hadamard Theorem that $\overline{\lim_{n\to\infty}}\sqrt[n]{|A_n|}=\frac 1 R$, and we are now to prove that $\overline{\lim_{k\to\infty}}\sqrt[k]{|c_k|}=\frac{1}{R-r}.$
But I have failed to prove this limit equality. Could you help me?