Power series are differentiable inside the radius of convergence

Statement

Let $f(x) = \sum_{n=0}^{\infty} a_n (x - c)^n$ be a power series with radius of convergence $R > 0$. Then $f$ is differentiable on $(c - R, c + R)$, and $$f'(x) = \sum_{n=1}^{\infty} n a_n (x - c)^{n-1}.$$ The derived series also has radius of convergence $R$.

Remarks

The Cauchy–Hadamard formula gives the radius of the derived series, since $\sqrt[n]{n} \to 1$. Then term-by-term differentiation applies on each $[c - r, c + r]$ with $r < R$. By induction, $f$ is infinitely differentiable, and $a_n = f^{(n)}(c)/n!$. So the series is the Taylor series of $f$ at $c$. Mathlib states differentiability as HasFPowerSeriesOnBall.differentiableOn.

Explanations

L1 human

A power series can be differentiated like a polynomial, one term at a time. The new series converges on the same open interval.