Cauchy-Hadamard formula

hbghlyj

math.drexel.edu/~tolya/Cauchy%20Hadamard%20formula.pdf

Theorem [Cauchy, 1821] The radius of convergence of the power series $\sum_{n=0}^{\infty} c_{n}\left(z-z_{0}\right)^{n}$ is
\[
R=\frac{1}{\overline{\lim }_{n \rightarrow \infty} \sqrt[n]{\left|c_{n}\right|}} .
\]
Example. For any increasing sequence of natural numbers $n_{j}$ the radius of convergence of the power series $\sum_{j=1}^{\infty} z^{n_{j}}$ is $R=1$.
Proof. Let $R=1 / \overline{\lim }_{n \rightarrow \infty} \sqrt[n]{\left|c_{n}\right|} \in[0, \infty]$.
If $R<\infty$, choose any $r>R$. Then $\varlimsup_{n \rightarrow \infty} \sqrt[n]{\left|c_{n}\right|}>1 / r$ and so $\left|c_{n}\right| r^{n}>1$ for infinitely many indices $n$. So $c_{n}\left(z-z_{0}\right)^{n}$ does not approach 0 for any $z$ with $\left|z-z_{0}\right|=r>R$. So the power series diverges for any $z$ with $\left|z-z_{0}\right|>R$.

If $R>0$, choose any $0<r<R$. Then $\overline{\lim }_{n \rightarrow \infty} \sqrt[n]{\left|c_{n}\right|}<1 / r$ and so $\left|c_{n}\right| r^{n}<1$ for all but finitely many indices $n$. Hence for any $z$ with $\left|z-z_{0}\right|<r$,
\[
\sum_{n=0}^{\infty}\left|c_{n}\right|\left|z-z_{0}\right|^{n}=\sum_{n=0}^{\infty}\left|c_{n}\right| r^{n}\left|\frac{z-z_{0}}{r}\right|^{n} \leq M \sum_{n=0}^{\infty}\left|\frac{z-z_{0}}{r}\right|^{n}<\infty .
\]
So the power series converges for any $z$ with $\left|z-z_{0}\right|<R$.
It follows that the radius of convergence is $R$.

hbghlyj

complex.pdf
Definition 3.15 (Radius of convergence). Let $\sum_{n=0}^{\infty} a_{n} z^{n}$ be a power series, and let $S$ be the set of $z \in \mathbb{C}$ at which it converges. The radius of convergence of the power series is $\sup \{|z|: z \in S\}$, or $\infty$ if the set $S$ is unbounded. Note that $S$ is always nonempty since $0 \in S$.

Proposition 3.16. Let $\sum_{n=0}^{\infty} a_{n} z^{n}$ be a power series, let $S$ be the subset of $\mathbb{C}$ on which it converges and let $R$ be its radius of convergence. Then we have
\[\tag{3.3}
B(0, R) \subseteq S \subseteq \bar{B}(0, R) .
\]
The series converges absolutely on $B(0, R)$ and if $0 \leq r<R$ then it converges uniformly on $\bar{B}(0, r)$. Moreover, we have
$$\tag{3.4}\frac{1}{R}=\limsup _{n}\left|a_{n}\right|^{1 / n}$$
Remark. The statement is uncontroversial when $0<R<\infty$. Suitably interpreted, the proposition makes sense when $R=0$ and $R=\infty$ as well, and we consider the statement to include these cases:
• When $R=0$, one should take $B(0, R)=\emptyset$ and $\bar{B}(0, R)=\{0\}$, so (3.3) is the statement that $S \subseteq\{0\}$ in this case (which is trivial). Statement (3.4) should be taken to mean that $\lim \sup _{n}\left|a_{n}\right|^{1 / n}=\infty$ (which is not so trivial).
• When $R=\infty$, one should take $B(0, R)=\bar{B}(0, R)=\mathbb{C}$, so (3.3) is the statement that $S=\mathbb{C}$. Statement (3.4) should be taken to mean that $\lim _{n \rightarrow \infty}\left|a_{n}\right|^{1 / n}=0$.
Proof. We begin with (3.3), which was essentially proven in Prelims. The containment $S \subseteq \bar{B}(0, R)$ is immediate from the definition of radius of convergence (even when $R=\infty$ ). The other containment $B(0, R) \subseteq S$, as well as the statement that the series converges absolutely on $B(0, R)$, are both consequences of the statement that the series converges uniformly on $\bar{B}(0, r)$ when $0 \leq r<R$. This is because $B(0, R)=\bigcup_{r<R} \bar{B}(0, r)$ (this is also true when $R=\infty)$.

Let us, then, prove this statement. By definition of $R$, there is some $w$, $|w|>r$, such that $\sum_{n=0}^{\infty} a_{n} w^{n}$ converges. In particular, the terms of the sum are bounded: $\left|a_{n} w^{n}\right| \leq M$ for some $M$. But then if $|z| \leq r$ we have
\[
\left|a_{n} z^{n}\right|=\left|a_{n} w^{n}\right|\left|\frac{z}{w}\right|^{n} \leq M\left|\frac{r}{w}\right|^{n} .
\]
The geometric series $\sum_{n}\left|\frac{r}{w}\right|^{n}$ converges, since $|w|>r$. Therefore, by the Weierstass test (for series) $\sum_{n=0}^{\infty} a_{n} z^{n}$ converges uniformly for $|z| \leq r$.

Now we turn to the formula (3.4), which is not always covered in Prelims. Suppose the radius of convergence is $R$. Let $0 \leq r<R$. By the above, there is some $w,|w|>r$, such that $\left|a_{n} w^{n}\right| \leq M$ for all $n$. We may clearly assume that $M \neq 0$. Taking $n$th roots gives $\left|a_{n}\right|^{1 / n}|w| \leq M^{1 / n}$. Since $M^{1 / n} \rightarrow 1$ as $n \rightarrow \infty$, this implies that $\lim \sup _n\left|a_n\right|^{1 / n} \leq \frac{1}{|w|}<\frac{1}{r}$. Since $r<R$ was arbitrary, it follows that $\lim \sup _n\left|a_n\right|^{1 / n} \leq \frac{1}{R}$. (This is perfectly legitimate when $R=\infty$ as well, with the interpretation that $\frac{1}{R}=0$ in this case.)

In the other direction, suppose that $\lim \sup _{n}\left|a_{n}\right|^{1 / n}=L$ and that $L \in$ $(0, \infty)$. If $L^{\prime}>L$, this means that $\left|a_{n}\right|^{1 / n} \leq L^{\prime}$ for all sufficiently large $n$. Therefore $\left|a_{n} z^{n}\right| \leq\left|L^{\prime} z\right|^{n}$ (for sufficiently large $n$ ), and by the geometric series formula the series $\sum_{n} a_{n} z^{n}$ converges provided $|z|<\frac{1}{L^{\prime}}$. Therefore $R \geq \frac{1}{L^{\prime}}$. Since $L^{\prime}>L$ was arbitrary, $R \geq \frac{1}{L}$, that is to say $\lim \sup _{n}\left|a_{n}\right|^{1 / n} \geq\frac1R$.

The argument is valid with minimal changes when $L=0$; we have shown that $R>\frac{1}{L^{\prime}}$ for all $L^{\prime}>0$, and so $R=\infty$, and so the inequality $\lim \sup _{n}\left|a_{n}\right|^{1 / n} \geq \frac{1}{R}$ remains true (with the interpretation discussed above).
When $L=\infty$, the inequality is vacuously true. Putting all this together concludes the proof.