# Complex power series are in $C^\infty$

I need to show that:

Let $$R>0$$ and let $$f:\{ z\in \mathbb{C} \mid \lvert z \rvert < R \} \rightarrow \mathbb{C}$$ be given by a convergent power series $$\sum_0^\infty a_n z^n$$. Show that $$f \in C^\infty$$.

I know the following facts:

Let the power series $$\sum_0^\infty a_n z^n$$ have convergence radius $$R>0$$. Then the function $$f:\{ z\in \mathbb{C} \mid \lvert z \rvert < R \} \rightarrow \mathbb{C}$$ given by $$\sum_0^\infty a_n z^n$$ is holomorphic and its derivative is $$\sum_{n=1}^\infty na_nz^{n-1}$$.

The convergence radius of $$\sum_{n=1}^\infty na_nz^{n-1}$$ is again $$R$$.

The following hint was given:

From real multivariate analysis we know that a continuous function, whose continuous first-order partial derivatives exist and are $$C^k$$, is also in $$C^{k+1}$$.

If I am not mistaken we can use the first two points to say that $$f^\prime$$ is holomorphic. If I am not mistaken, we can repeatedly use this argumentation to prove that $$f \in C^\infty$$; so I do not understand why this hint was given. Could you explain?

• It is not entirely clear what you are supposed to do. The second yellow box states that $f$ is differentiable and $f'$ has the same radius of convergence. Then applying the result to $f'$ and using induction it follows that it is $C^\infty$. Feb 5, 2020 at 18:17

The radius of convergence $$R$$ of $$\sum a_n z^n$$ is given by

$$\frac{1}{R}= \lim \sup |a_n^{1/n}|.$$

So, the radius of convergence $$R^\prime$$ of $$\sum n a_n z_n$$ is

$$\frac{1}{R^\prime}= \lim \sup | (n a_n)^{1/n}|= \lim \sup | n^{1/n} a_n^{1/n}|=\frac{1}{R}$$

because $$\lim_{n\rightarrow \infty} n^{1/n} \rightarrow 1.$$

Let $$q=n^{1/n}$$. Then $$\lim_{n\rightarrow \infty } \log q =\frac{\log n}{n} =0$$ by L'Hopital's rule and thus $$q \rightarrow 1.$$

For the last part ($$f \in C^\infty$$):

We have just shown that $$f$$ analytic in a disk of radius $$R \Rightarrow f^\prime$$ is analytic in the same disk of radius $$R$$. So $$f^{k}$$ analytic $$\Rightarrow f^{k+1}$$ analytic. By induction $$f^{n}$$ is analytic for all $$n$$. Thus $$f\in C^\infty$$.