[proofplan] We first show that multiplying the coefficients by $n$ and shifting the index does not change the [radius of convergence](/theorems/262), using the root-test characterization of the radius. Then, around an arbitrary point $w \in B(z_0,R)$, we choose a smaller closed disc on which both the original series and a slightly stronger weighted series converge absolutely and uniformly. This uniform control justifies passing the limit through the difference quotient and gives the termwise derivative formula. Since $w$ was arbitrary, the formula holds throughout the disc. [/proofplan] [step:Show that the differentiated series has the same radius of convergence] Let \begin{align*} L := \limsup_{n \to \infty} |a_n|^{1/n}. \end{align*} By the root-test characterization of the [radius of convergence](/theorems/265) of a [power series](/page/Power%20Series), the [radius of convergence](/theorems/273) of the original power series is $R = L^{-1}$, where $0^{-1}=\infty$ and $\infty^{-1}=0$. Since $R>0$, we have $L<\infty$. Define the coefficient sequence $(b_m)_{m=0}^{\infty}$ of the differentiated power series by \begin{align*} b_m := (m+1)a_{m+1}. \end{align*} Then the differentiated series can be written as \begin{align*} \sum_{m=0}^{\infty} b_m(z-z_0)^m. \end{align*} Its reciprocal radius is \begin{align*} \limsup_{m \to \infty} |b_m|^{1/m} = \limsup_{m \to \infty} (m+1)^{1/m}|a_{m+1}|^{1/m}. \end{align*} Because $(m+1)^{1/m} \to 1$ and \begin{align*} |a_{m+1}|^{1/m} = \left(|a_{m+1}|^{1/(m+1)}\right)^{(m+1)/m}, \end{align*} the finite-limsup sequence $|a_{m+1}|^{1/(m+1)}$ has the same limsup after raising to the powers $(m+1)/m \to 1$. Deleting the initial coefficient does not change the limsup, so \begin{align*} \limsup_{m \to \infty} |b_m|^{1/m} = \limsup_{m \to \infty} |a_{m+1}|^{1/(m+1)} = L. \end{align*} Thus the differentiated series has radius $L^{-1}=R$. [guided] The radius of convergence is controlled by the exponential growth rate of the coefficients. Multiplying the $n$-th coefficient by $n$ should not affect that exponential growth rate, because $n^{1/n} \to 1$. We make this precise. Define \begin{align*} L := \limsup_{n \to \infty} |a_n|^{1/n}. \end{align*} By the root-test characterization of the radius of convergence, the original radius is $R=L^{-1}$, with $0^{-1}=\infty$. The differentiated series has coefficients \begin{align*} b_m := (m+1)a_{m+1}, \end{align*} because \begin{align*} \sum_{n=1}^{\infty} n a_n(z-z_0)^{n-1} = \sum_{m=0}^{\infty} (m+1)a_{m+1}(z-z_0)^m. \end{align*} Its reciprocal radius is therefore \begin{align*} \limsup_{m \to \infty} |b_m|^{1/m} = \limsup_{m \to \infty} (m+1)^{1/m}|a_{m+1}|^{1/m}. \end{align*} The factor $(m+1)^{1/m}$ tends to $1$, so it does not change the limsup. The remaining exponent differs from the natural exponent only by a factor tending to $1$: \begin{align*} |a_{m+1}|^{1/m} = \left(|a_{m+1}|^{1/(m+1)}\right)^{(m+1)/m}. \end{align*} Since $L<\infty$ and $(m+1)/m \to 1$, raising the non-negative sequence $|a_{m+1}|^{1/(m+1)}$ to these exponents preserves its limsup. Also, passing from $n$ to $m+1$ deletes only the initial coefficient, so \begin{align*} \limsup_{m \to \infty} |a_{m+1}|^{1/(m+1)} = L. \end{align*} Hence \begin{align*} \limsup_{m \to \infty} |b_m|^{1/m} = \limsup_{m \to \infty} |a_{m+1}|^{1/(m+1)} = L. \end{align*} The differentiated series therefore has radius $L^{-1}$, which is exactly the original radius $R$. [/guided] [/step] [step:Choose a local closed disc with uniform summability] Fix $w \in B(z_0,R)$. Choose [real numbers](/page/Real%20Numbers) $\rho$ and $\sigma$ such that \begin{align*} |w-z_0| < \rho < \sigma < R. \end{align*} If $R=\infty$, choose any finite $\sigma>\rho$. Since the original power series has radius $R$, the numerical series \begin{align*} \sum_{n=0}^{\infty} |a_n|\sigma^n \end{align*} converges. For every $n \ge 1$, \begin{align*} n |a_n|\rho^{n-1} = \frac{n}{\rho}\left(\frac{\rho}{\sigma}\right)^n |a_n|\sigma^n. \end{align*} The sequence \begin{align*} n\left(\frac{\rho}{\sigma}\right)^n \end{align*} is bounded because $0<\rho/\sigma<1$. Therefore there is a constant $M_{\rho,\sigma}>0$ such that \begin{align*} n |a_n|\rho^{n-1} \leq M_{\rho,\sigma}|a_n|\sigma^n \end{align*} for every $n \ge 1$. Hence \begin{align*} \sum_{n=1}^{\infty} n |a_n|\rho^{n-1} \end{align*} converges. [/step] [step:Compute the difference quotient under the summation sign] Let $x := w-z_0$. Choose $\delta>0$ such that \begin{align*} |x|+\delta \leq \rho. \end{align*} For $h \in \mathbb{C}$ with $0<|h|<\delta$, define \begin{align*} Q_h := \frac{f(w+h)-f(w)}{h}. \end{align*} Since $|x+h|\leq \rho<R$, both $w$ and $w+h$ lie in $B(z_0,R)$, and the two power series defining $f(w+h)$ and $f(w)$ converge absolutely. Therefore subtraction may be performed term by term as the limit of the corresponding finite partial sums. For each integer $n \ge 1$, the finite algebraic identity \begin{align*} \frac{(x+h)^n-x^n}{h} = \sum_{k=0}^{n-1}(x+h)^k x^{n-1-k} \end{align*} holds. Thus \begin{align*} Q_h = \sum_{n=1}^{\infty} a_n \sum_{k=0}^{n-1}(x+h)^k x^{n-1-k}. \end{align*} Moreover, \begin{align*} \left|a_n \sum_{k=0}^{n-1}(x+h)^k x^{n-1-k}\right| \leq n|a_n|\rho^{n-1}. \end{align*} The dominating numerical series from the previous step converges, so the Weierstrass majorant test gives absolute and [uniform convergence](/page/Uniform%20Convergence) of these difference-quotient series for $0<|h|<\delta$. For each fixed $n \ge 1$, \begin{align*} \lim_{h \to 0} a_n \sum_{k=0}^{n-1}(x+h)^k x^{n-1-k} = n a_n x^{n-1}. \end{align*} Since the summable majorant $n|a_n|\rho^{n-1}$ is independent of $h$, the [dominated convergence theorem](/theorems/4) for series applies to the counting measure on $\mathbb{N}$. Passing the limit through the series gives \begin{align*} \lim_{h \to 0} Q_h = \sum_{n=1}^{\infty} n a_n x^{n-1} = \sum_{n=1}^{\infty} n a_n(w-z_0)^{n-1}. \end{align*} [/step] [step:Conclude differentiability at every point of the disc] The previous step proves that the complex difference quotient \begin{align*} \frac{f(w+h)-f(w)}{h} \end{align*} has a limit as $h \to 0$ through nonzero complex numbers, and that the limit is \begin{align*} \sum_{n=1}^{\infty} n a_n(w-z_0)^{n-1}. \end{align*} Therefore $f$ is complex differentiable at $w$, and \begin{align*} f'(w)=\sum_{n=1}^{\infty} n a_n(w-z_0)^{n-1}. \end{align*} Since $w \in B(z_0,R)$ was arbitrary, $f$ is complex differentiable on $B(z_0,R)$ and the displayed derivative formula holds for every $z \in B(z_0,R)$. Together with the first step, this proves both the termwise differentiation formula and equality of the radii of convergence. [/step]