---
tag: 0YMP521ZR5
kind: statement
role: theorem
name: Power series are holomorphic in the disc of convergence
aliases: [termwise-differentiation-of-power-series]
msc: [30B10]
uses: [0VARVQ9ZW1, 045EWAXY9P, 0DG0Z72CQD]
sources:
  - url: https://en.wikipedia.org/wiki/Power_series
    locator: Differentiation and integration
  - url: https://en.wikipedia.org/wiki/Power_series
    locator: Analytic functions
---

## Statement

Let $\sum_{n \ge 0} a_n (z - c)^n$ be a [[power series|0VARVQ9ZW1]] with radius of convergence $R > 0$, and let $D = \{z : |z - c| < R\}$ be its [[disc of convergence|045EWAXY9P]].
Let $f(z) = \sum_{n=0}^{\infty} a_n (z - c)^n$ for $z \in D$.
Then $f$ is [[holomorphic|0DG0Z72CQD]] on $D$, and for each $z \in D$,
$$f'(z) = \sum_{n=1}^{\infty} n a_n (z - c)^{n-1}.$$
The series for $f'$ has the same radius of convergence $R$.

## Remarks

By induction, $f$ has complex derivatives of all orders on $D$, and $a_n = f^{(n)}(c) / n!$. So the power series is the Taylor series of $f$ at $c$.
For example, the [[complex exponential|054VXHMCA3]] has $R = \infty$, so it is [[entire|0G2YJN0H5S]].
The converse is also true: [[holomorphic functions are analytic|0826SGV3R0]].

## Explanations

### L1 human

Inside its disc of convergence, a power series can be differentiated term by term, like a polynomial.
The result is again a power series with the same disc, so you can do it again and again.