12.2 Radius of Convergence

12.8   Theorem. Let $\sum\{a_nz^n\}$ be a power series. Suppose $\sum\{a_nw^n\}$ converges for some $w\in\mbox{{\bf C}}\backslash\{0\}$. Then $\sum\{a_nz^n\}$ converges absolutely for all $z\in D(0,\vert w\vert)$.

Proof: Since $\sum\{a_nw^n\}$ converges, $\{a_nw^n\}$ is a null sequence, and hence is bounded. Say $\vert a_nw^n\vert\leq M$ for all $n\in\mbox{{\bf N}}$. Let $z\in D(0,\vert w\vert)$, so $\vert z\vert<\vert w\vert$, and let $${R= {{\vert z\vert}\over {\vert w\vert}}<1}$$. Then for all $n\in\mbox{{\bf N}}$

$$\vert a_nz^n\vert=\vert a_nw^n\vert \left\vert {z\over w}\right\vert^n\leq MR^n.$$

Now $\sum\{MR^n\}$ is a convergent geometric series, so by the comparison test, $\sum\{\vert a_nz^n\vert\}$ converges; i.e., $\sum\{a_nz^n\}$ is absolutely convergent. $\mid\!\mid\!\mid$

12.9   Corollary. Let $\sum\{a_nz^n\}$ be a power series. Suppose $\sum\{a_nw^n\}$ diverges for some $w\in\mbox{{\bf C}}$. Then $\sum\{a_nz^n\}$ diverges for all $z\in\mbox{{\bf C}}$ with $\vert z\vert>\vert w\vert$.

Proof: Suppose $\vert z\vert>\vert w\vert$. If $\sum\{a_nz^n\}$ converges, then by the theorem, $\sum\{a_nw^n\}$ would also converge, contrary to our assumption. $\mid\!\mid\!\mid$

12.10   Theorem. Let $\sum\{c_nz^n\}$ be a power series. Then one of the following three conditions holds:

a) $\sum\{c_nz^n\}$ converges only when $z=0$.

b) $\sum\{c_nz^n\}$ converges for all $z\in\mbox{{\bf C}}$.

c) There is a number $R\in\mbox{${\mbox{{\bf R}}}^{+}$}$ such that $\sum\{c_nz^n\}$ converges absolutely for $\vert z\vert<R$ and diverges for $\vert z\vert>R$.

Proof: Suppose that neither a) nor b) is true. Then there are numbers $w,v\in\mbox{{\bf C}}\backslash\{0\}$ such that $\sum\{c_nw^n\}$ converges and $\sum\{c_nv^n\}$ diverges. If $${a={{\vert w\vert}\over 2}}$$, and $b=2\vert v\vert$, it follows that $\sum\{c_na^n\}$ converges and $\sum\{c_nb^n\}$ diverges. By a familiar procedure, build a binary search sequence $\{[a_k,b_k]\}$ such that $[a_0,b_0]$ $=[a,b]$, and for all $k \in\mbox{{\bf N}}$, $\sum\{c_na_k^n\}$ converges and $\sum\{c_kb_k^n\}$ diverges. Let $R$ be the number such that $\{[a_k,b_k]\}\to R$. Then $a_k\leq R\leq b_k$ for all $k \in\mbox{{\bf N}}$ and $\lim\{a_k\}=\lim\{b_k\}=R$. If $\vert z\vert<R$, then for some $k \in\mbox{{\bf N}}$ we have $\vert a_k-R\vert<R-\vert z\vert$, and

$$\begin{eqnarray*} \vert a_k-R\vert<R-\vert z\vert &\mbox{$\Longrightarrow$}&a_k>... ... \\ &\mbox{$\Longrightarrow$}&\sum\{c_nz^k\}\mbox{ converges. } \end{eqnarray*}$$

If $\vert z\vert>R$, then for some $k \in\mbox{{\bf N}}$ we have $\vert b_k-R\vert<\vert z\vert-R$, and

$$\begin{eqnarray*} \vert b_k-R\vert<\vert z\vert-R&\mbox{$\Longrightarrow$}&b_k <... ...row$}&\sum\{c_nz^n\}\mbox{ diverges. }\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

12.11   Definition (Radius of convergence.) Let $\{\sum c_nz^n\}$ be a power series. If there is a number $R\in\mbox{${\mbox{{\bf R}}}^{+}$}$ such that $\sum\{c_nz^n\}$ converges for $\vert z\vert<R$, and diverges for $\vert z\vert>R$, we call $R$ the radius of convergence of $\sum\{c_nz^n\}$. If $\sum\{c_nz^n\}$ converges only for $z=0$, we say $\sum\{c_nz^n\}$ has radius of convergence $0$. If $\sum\{c_nz^n\}$ converges for all $z\in\mbox{{\bf C}}$, we say $\sum\{c_nz^n\}$ has radius of convergence $\infty$.

If a power series has radius of convergence $R\in\mbox{${\mbox{{\bf R}}}^{+}$}$, I call $D(0,R)$ the disc of convergence for the series, and I call $C(0,R)$ the circle of convergence for the series. If $R=\infty$, I call $\mbox{{\bf C}}$ the disc of convergence of the series (even though $\mbox{{\bf C}}$ is not a disc).

12.12   Example. I will find the radius of convergence for $${\sum\left\{ {(3n)! z^n \over n!(2n)!}\right\}}$$. I will apply the ratio test. Since the ratio test applies to positive sequences, I will consider absolute convergence. Let $${a_n = {(3n)!z^n \over n! (2n)!}}$$ for all $n\in\mbox{{\bf N}}$. Then for all $z\in\mbox{{\bf C}}\setminus \{0\}$,

$$\begin{eqnarray*} {\vert a_{n+1}\vert\over\vert a_n\vert} &=& { (3(n+1))! \vert ... ... \over (1+{1\over n})(2+{1\over n})(2+{2\over n})} \vert z\vert. \end{eqnarray*}$$

Hence

$$\left\{{\vert a_{n+1}\vert \over \vert a_n\vert}\right\} \to ... ...\over 1\cdot 2 \cdot 2}\vert z\vert = {27\vert z\vert \over 4}.$$

By the ratio test, $\sum\{a_n\}$ is absolutely convergent if $\vert z\vert < {4\over 27}$, and is divergent if $\vert z\vert > {4\over 27}$. It follows that the radius of convergence for our series is ${4\over 27}$.

12.13   Exercise. Find the radius of convergence for the following power series:

a)

$${ \sum \{ 3^n \sqrt{n} z^n\}_{n\geq 1}}$$.

b)

$${ \sum\left\{ {z^n \over n^n}\right\}_{n\geq 1}}$$.

12.14   Exercise. A Let $r$ be a positive real number.

a)

Find a power series whose radius of convergence is equal to $r$.

b)

Find a power series whose radius of convergence is $\infty$.

c)

Find a power series whose radius of convergence is $0$.