Expansion of a series in a point different from its center. I want to prove the following:

let $f$ be a power series centered at the origin. Prove that $f$ has a power series expansion around any point in its disc of convergence.

What I did:
$$\displaystyle\sum_{n=0}^\infty a_nz^n= \displaystyle\sum_{n=0}^\infty a_n(z_0+(z-z_0))^n=\displaystyle\sum_{n=0}^\infty a_n\displaystyle\sum_{k=0}^n \binom{n}{k} z_0^k(z-z_0)^{n-k}$$
And if we consider the partial sum of this series we find the closed form:
$$\displaystyle\sum_{n=0}^k a_nz^n=\displaystyle\sum_{n=0}^k\left(\displaystyle\sum_{s=n}^k\binom{s}{n} a_s z_0^s \right)(z-z_0)^n $$
Now, we can verify that the coefficients converge since
$$\displaystyle\sum_{s=n}^\infty |a_s|\binom{s}{n}|z_0|^s\leq \displaystyle\sum_{s=n}^\infty |a_s|\displaystyle\frac{s!}{(s-n)!}|z_0|^s $$
Now I want to know if from this it follows that 
$$\displaystyle\sum_{n=0}^\infty a_nz^n=\displaystyle\sum_{n=0}^\infty\left(\displaystyle\sum_{s=n}^\infty\binom{s}{n} a_s z_0^s \right)(z-z_0)^n $$
how do I prove that the convergence radio is  $R-|z_0|$ where $R$ is the convergence radius of the expansion in $z=0$?
Thanks!
 A: In addition to Robert Israel's answer, may I point out that there is also an alternative way to approach this?
If $f$ is a power series, centred around $0$ with radius of convergence $R$, then $f$ is certainly holomorphic in $B(0,R)$, because any power series can be differentiated term by term.
Now, suppose $z_0$ is another point in $B(0,R)$, and suppose $C(z_0, R')$ is a circle around $z_0$ contained inside $B(0,R)$. Then for any $z \in B(z_0, R')$, you can write
\begin{align*} f(z) & = \frac 1 {2\pi i}\oint_{C(z_0, R')} \frac {f(w)}{w - z}  dw \\ & =  \frac 1 {2\pi i}\oint_{C(z_0, R')} \frac {f(w) }{(w - z_0) - (z - z_0)} dw \\ & =   \frac 1 {2\pi i}\oint_{C(z_0, R')} f(w) \sum_{n = 0}^{\infty} \frac {(z - z_0)^n}{(w - z_0)^{n+1}} dw\\ & =   \sum_{n = 0}^{\infty} \left( \frac 1 {2\pi i}\oint_{C(z_0, R')} \frac { f(w) }{(w - z_0)^{n+1}} dw \right)  (z - z_0)^n \end{align*}
[It is legitimate to move the sum outside the integral in the final line because the geometric series is uniformly convergent on the circle.]
But look! We've explicitly written $f$ as a Taylor series around $z_0$! And that's it.
A: The convergence radius is at least $R - |z_0|$, because the double series converges absolutely for $|z - z_0| < R - |z_0|$.  It can be greater (e.g. look at some simple examples such as the series for $1/(1-z)$). 
