Power series is infinitely differentiable in disk of convergence

Statement

Consider the power series about a point $z_0\in \mathbb{C}$ with coefficients $a_n\in \mathbb{C}$:

$\sum a_n(z-z_0{)}^n$

If $R$ denotes the radius of convergence of this power series, then the function given by:

$f(z)=\sum a_n(z-z_0{)}^n$

is infinitely differentiable at all $z$ with $|z-z_0|<R$. Further, the derivative at any point $z$ in this disk is given by:

$f^{\prime }(z)=\sum _{n=1}^{\mathrm{\infty }}n{a}_n(z-z_0{)}^{n-1}$.

In particular:

• When $R=0$, the function is not defined around $z_0$, so the statement says nothing.
• When $0<R<\mathrm{\infty }$, the function $f$ is infinitely differentiable at all points in an open disk around $z_0$. In particular, it is infinitely differentiable at $z_0$.
• When $R=\mathrm{\infty }$, $f$ is infinitely differentiable at all points.

Definitions used

Radius of convergence

Further information: Radius of convergence

Proof

Same radius of convergence

We first argue that the power series for $f$ and for the purported $f^{\prime }$ have the same radius of convergence.

The radius of convergence of the purported $f^{\prime }$ is given by:

$\frac{1}{lim{sup}_{n\to \mathrm{\infty }}{\left|n{a}_n\right|}^{1/(n-1)}}$

The $n^{1/(n-1)}$ part does not contribute to the limit, since it tends to $1$. Further, we see that if $|a_n{|}^{1/n}\to \mathrm{\infty }$, so does $|a_n{|}^{1/(n-1)}$, and if $|a_n{|}^{1/n}\to 0$, so does $|a_n{|}^{1/(n-1)}$. In case either goes to a finite limit, the other goes to the same limit. Thus, the radii of converge are the same, and we are done.

Proof of once-differentiability, and power series for derivative function

Given: $z$ such that $|z-z_0|<R$.

To prove: The expression:

$\underset{\left|h\right|\to 0}{lim}\frac{f(z+h)-f(z)}{h}$

is well-defined and equals:

$\sum _{n=1}^{\mathrm{\infty }}n{a}_n(z-z_0{)}^{n-1}$.

Proof: For $h$ sufficiently small, $z+h$ is also in the disk of convergence, so the power series for $f(z+h)$ and for $f(z)$ both converge absolutely. Thus, their difference also converges absolutely. Rearranging this absolutely convergent series, we get:

$\underset{\left|h\right|\to 0}{lim}\frac{f(z+h)-f(z)}{h}=\underset{\left|h\right|\to 0}{lim}\sum _{n=1}^{\mathrm{\infty }}\frac{a_n(z+h-z_0{)}^n-(z-z_0{)}^n}{h}$

Now, we already argued that the radius of convergence of the function:

$\sum _{n=1}^{\mathrm{\infty }}n{a}_n(z-z_0{)}^{n-1}$

is equal to $R$. Now, this function can be rewritten as:

$\sum _{n=1}^{\mathrm{\infty }}\underset{\left|h\right|\to 0}{lim}\frac{a_n(z+h-z_0{)}^n-(z-z_0{)}^n}{h}$

The completion of the proof thus rests on justifying exchange of limit and summation.

Proof of infinite differentiability

To prove infinite differentiability, we use the fact that the radius of convergence of the derivative of $f$ is the same as the radius of convergence of $f$, to iterate the argument with $f^{\prime }$.