Proving a convergent series defines an analytic function Assume $ f\left(z\right)=\sum_{n=0}^{\infty}a_{n}z^{n} $ converges on $D(0,r)$ which is a disk centered at the origin with radius $ r $.
How can I prove that $ f $ can be described as $ f\left(z\right)=\sum_{n=0}^{\infty}b_{n}\left(z-z_{0}\right)^{n} $ for any $ z_0 \in D$ and with the maximal radius $r'$ such that $ D\left(z_{0},r'\right)\subset D\left(0,r\right) $?
What I thought is to show the obvious, using $ b_{n}=\frac{f^{(n)}\left(z_{0}\right)}{n!} $ and then to show that the reminder in the form of lagrange tends to $0 $. But, we have never proved lagrange reminder theorem for complex function so I feel uncomfortable with that, and besides, I feel like there's much more efficient way.
Any help would be appreciated. Thanks in advance.
(also, I'd prefer a way without line integrals, if exists)
 A: The direct proof of the result above has three steps which I will outline as the computations/profs are standard
Step 1: using the usual results about expressing the radius of convergence in terms of coefficients,  the fact that a power series is uniformly and absolutely convergent on $|z| \le r_1 <r$ for any $r_1<r$, and the standard theorem about differentiating series of functions, one gets that for every $m \ge 1, f^{(m)}(z_0)$ is computable by differentiating the series $\sum_{n=0}^{\infty}a_{n}z^{n}$ term by term, so it is
$f^{(m)}(z_0)=\sum_{n \ge m}n(n-1)...(n-m+1)a_nz_0^{n-m}$
Also one has $\sum_{n \ge 0}|a_n||w|^n < \infty, |w|<r$
Step 2: By the triangle inequality and the usual expansion of $(z_0+z)^n$, the double series $f(z_0+z)=\sum_{n=0}^{\infty}a_{n}(z_0+z)^{n}=\sum_{n \ge 0}(a_n\sum_{m=0}^nn(n-1)...(n-m+1)/m!a_nz_0^{n-m}z^m)$ is absolutely convergent for $|z_0|+|z|<r$ and uniformly convergent for $|z_0|+|z| \le r_1<r$ for any $r_1<r$, so can be rearranged at will, since absolute convergence reduces to the fact that
$\sum_{n \ge 0}|a_n|(|z_0|+|z|)^n < \infty$ for $|z_0|+|z|<r$ as in Step 1 above with $w=|z_o|+|z|<r$
Step 3: Changing the order of summation in the double series above and using Step 1, one immediately gets that $f(z_0+z)=\sum_{m \ge 0}f^{(m)}(z_0)z^m/m!$ as long as $|z_0|+|z|<r$ which is the required result substituting $z \to z-z_0$
