How can I rewrite $\sum_{n=0}^{\infty}a_n\sum_{k=0}^{n}\binom{n}{k}(z-z_0)^kz_0^{n}$ to $\sum_{n=0}^{\infty}b_n(z-z_0)^{n}$?

Question

Unfortunately I don't know how to proceed, we had the excercise that we should show if $P(z)=\sum_{n=0}^{\infty}a_nz^n$ converges for $z_0$ one can find a powerseries around this Point $P_{z_0}(z)= \sum_{n=0}^{\infty}b_n(z-z_0)^{n}$ such that $P_{z_0}(z)=P(z)$, $\forall z\in D_{R-|z_0|}(z_0)$ where $R$ is the radius of convergence of $P(z)$ and $ D_{R-|z_0|}(z_0)$ is the open disk around $z_0$ with the radius $R-|z_0|$.

A solution or a hint would be welcommed.

Answer 1

$D_{R-|z_0|} (z_0) \subset D_R(0)$. Since $P$ is analytic in $D_R(0)$ it is analytic on $D_{R-|z_0|} (R)$. Hence it has a power series expansion around $z_0$ in that disk.