# Statements > [!NOTE] Theorem > Let $\sum_{n=0}^{\infty} a_{n}z^{n}$ be a power series with [[Radius of Convergence of Complex Power Series|radius of convergence]] $R>0$. Then for $\lvert z \rvert< R$ the function $f: z\mapsto \sum_{n\geq 0 } a_{n} z^{n}$ is [[Complex Differentiability|complex differentiable]] and $f'(z) = \sum_{n=1}^{\infty} na_{n} z^{n-1}.$ **Remark**: It's pointless to add the case $R=0$. In this case, the power series is not convergent for all non-zero values of $z$. > [!NOTE] Theorem (The differentiability of power series II) > Let $f(x)=\sum a_{n} x^{n}$ be a *power series* with *radius of convergence* $R$. Then $f$ is [[Fréchet Differentiation|differentiable]] on $(-R,R)$ and $f'(x)=\sum_{n=1}^{\infty} n a_{n} x^{n-1}$ # Proofs ###### Proof (Analysis 3) First note that $f'$ as defined above has the same radius of convergence as $f$ since $\lim \sup \lvert na_{n} \rvert^{1/n} = \lim n^{1/n} \lim \sup \lvert a_{n} \rvert^{1/n} = \lim \sup \lvert a_{n} \rvert^{1/n}. $ We now consider $\frac{f(w)-f(z)}{w-z} - \sum_{n=1}^{\infty} na_{n} z^{n-1} = \sum_{n=0} a_{n} \left( \frac{w^n -z^n}{w-z} - nz^{n-1}\right). \tag{1}$ Notice that $\frac{w^n - z^n}{w-z} = w^{n-1} + w^{n-2} z + \cdots+ wz^{n-2}+ z^{n-1}.$ ###### Proof (Analysis 2) WTP the following [[Limit of Real Function at a Point|limit]] $\frac{ \sum_{n=0}^{\infty} a_{n} y^{n} - \sum_{n=0}^{\infty} a_{n} x^{n} }{y-x} = \sum_{n=0}^{\infty} a_{n} \frac{y^{n}-x^{n}}{y-x} \to \sum_{n=1}^{\infty} n a_{n} x^{n-1}$as $y\to x$. First observe that $\begin{align}& \left\lvert \sum_{n=0}^{\infty} a_{n} \frac{y^{n}-x^{n}}{y-x} - \sum_{n=1}^{\infty} na_{n}x^{n-1} \right\rvert \\ &\leq \left\lvert \sum_{n=N+1}^{\infty} a_{n} \frac{y^{n}-x^{n}}{y-x} \right\rvert + \left\lvert \sum_{n=0}^{N} a_{n} \frac{y^{n}-x^{n}}{y-x} - \sum_{n=1}^{N} na_{n}x^{n-1} \right\rvert + \left\lvert \sum_{n=N+1}^{\infty} na_{n}x^{n-1} \right\rvert. \end{align}$ For any $x\in (-R,R)$, fix $T$ with $|x|<T<R$. By [[Radius of Convergence of Derivative of Real Power Series About Zero]], $\sum n|a_{n}|T^{n-1}$ converges so given $\varepsilon>0$ there is a number $N$ so that $\sum_{n=N+1}^{\infty} n |a_{n}|T^{n-1} < \frac{\varepsilon}{3}.$Now if $0<|y-x|<\delta_{1}:=T-|x|$ we have $|y|<T$ and $|x|<T$ so $\left\lvert \sum_{n=N+1}^{\infty} na_{n}x^{n-1} \right\rvert \leq \sum_{n=N+1}^{\infty} n |a_{n}| |x|^{n-1} < \frac{\varepsilon}{3} $and also $\begin{align} \left\lvert \sum_{n=N+1}^{\infty} a_{n} \frac{y^{n} - x^{n}}{y-x} \right\rvert &= \left\lvert \sum_{n=N+1}^{\infty} a_{n} (y^{n-1}+ y^{n-2} x +\dots + x^{n-1} ) \right\rvert \\ & \leq \sum_{n=N+1}^{\infty} |a_{n}| (|y|^{n-1} +\dots + |x|^{n-1} ) \\ & \leq \sum_{n=N+1}^{\infty} n |a_{n}|T^{n-1} < \frac{\varepsilon}{3} \end{align}$ Next, the sum $\sum_{n=1}^{N} a_{n} (y^{n-1} + y^{n-2} +\dots+x^{n-1} )$is a [[Ring of Polynomial Forms|polynomial]] in $y$ (which must be cts) whose values at $x$ is $\sum_{1}^{N} na_{n}x^{n-1}$ so $\exists\delta_{2}>0$ such that if $0<|y-x|<\delta_{2}$ then $\begin{align} &\left\lvert \sum_{n=0}^{N} a_{n} \frac{y^{n}-x^{n}}{y-x} - \sum_{n=1}^{N} na_{n}x^{n-1} \right\rvert \\ & = \left\lvert \sum_{n=1}^{N} a_{n} (y^{n-1} + y^{n-2} +\dots+x^{n-1} ) - \sum_{n=1}^{N} na_{n}x^{n-1} \right\rvert < \frac{\varepsilon}{3}. \end{align}$So finally if $0<|y-x|<\delta:= \min(\delta_{1},\delta_{2})$ then $\begin{align}& \left\lvert \sum_{n=0}^{\infty} a_{n} \frac{y^{n}-x^{n}}{y-x} - \sum_{n=1}^{\infty} na_{n}x^{n-1} \right\rvert \\ &\leq \left\lvert \sum_{n=N+1}^{\infty} a_{n} \frac{y^{n}-x^{n}}{y-x} \right\rvert + \left\lvert \sum_{n=0}^{N} a_{n} \frac{y^{n}-x^{n}}{y-x} - \sum_{n=1}^{N} na_{n}x^{n-1} \right\rvert + \left\lvert \sum_{n=N+1}^{\infty} na_{n}x^{n-1} \right\rvert < \varepsilon . \end{align} $