# Proving that $\sum_{k=0}^{\infty} \frac{(\partial^{k}f/\partial_{z}(P))}{k!}(z-P)^{k}$ converges?

In the text "Function Theory of One Complex Variable Third Edition, how did the author come to the conclusions in $(*)$, and also how would you justify that each of the integrals are equivalent to each other ?

$\text{Theorem (3.3.1)}$

Let $U \subset \mathbb{C}$ be an open subset and let $f$ be holomorphic on $U$ Let $P \in U$ and suppose that $D(P,r) \subset \mathbb{U}$ then

$$\sum_{k=0}^{\infty} \frac{(\partial^{k}f/\partial_{z}(P))}{k!}(z-P)^{k}$$

has a radius of convergence of at least $r$. It converges to $f(z)$ on $D(P,r)$

$\text{Proof}$

The author recalls that a key consequence of the Cauchy Differentiation Formula is that $f$ is $C^{\infty}$,and also one has an arbitrary $z \in D(P,r)$ the author initially proves the convergence of the series at this arbitrary $z$.

$\text{Lemma (1)}$

Let $r'$ be a positive number greater than $|z-P|$ but less than $r$ such that:

$$z \in D(P,r') \subset \overline D(P,r') \subset D(P,r)$$

$\text{Interjection}$

It seems by letting $|z-P| \leq r'$ such that $z \in D(P,r') \subset \overline D(P,r') \subset D(P,r)$ allows for the author to apply the Cauchy Integral Formula later on in the proof

$\text{Lemma (2)}$

Thus for $z \in D(P,r')=D(0,r')$ we have in $(*)$

$(*)$

$$f(z) = \frac{1}{2 \pi i} \oint_{|\zeta|=r'} \frac{f(\zeta)}{\zeta - z} = \frac{1}{2 \pi i} \oint_{|\zeta|=r'} \frac{f(\zeta)}{\zeta} \frac{1}{1 - z \cdot \zeta^{-1}}d \zeta = \frac{1}{2 \pi i } \oint_{|\zeta|=r'} \frac{f(\zeta)}{\zeta}\sum_{k=0}^{\infty}(z \cdot \zeta^{-1}) d \zeta.$$

The next equality uses simply $$\zeta - z = \zeta(1 - z\cdot\zeta^{-1})$$.
The interesting bit is the last equality, where the formula for the sum of a geometric series is used: $$\frac1{1 - z\cdot\zeta^{-1}} = \sum_{k=0}^{\infty}(z\cdot\zeta^{-1})$$ Now, by the uniform convergence of the series in $\{\zeta\in\Bbb C:|\zeta|=r'\}$, $\oint$ and $\sum$ can be exchanged...