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Math412 Notes Select Topics in Mathematical Analysis of Functions of One or Several Variables

Zheng-Chao Han

Section 3.1 Definition of a Power Series and Its Radius of Convergence

Definition 3.1.1. Definition of a Power Series.

A power series centered at \(x_{0}\) is a series of the form
\begin{equation*} a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})^{2}+\cdots =\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n} \end{equation*}
for some coefficients \(a_{0}, a_{1}, a_{2}, \cdots\text{.}\)
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A basic property of a power series is the following
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Proof.

Under the assumption that the power series \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\) converges at \(y\ne x_{0}\text{,}\) we know that \(a_{n}(y-x_{0})^{n} \to 0\) as \(n\to \infty\text{.}\) Thus there exists some \(C\gt 0\) such that \(\vert a_{n}(y-x_{0})^{n}\vert \le C\) for all \(n\text{.}\)
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For any \(z\) with \(|z-x_{0}| \lt |y-x_{0}|\text{,}\) we can find some \(0 \lt r \lt |y-x_{0}|\) such that \(|z-x_{0}|\le r \text{.}\) In fact, for any \(0 \lt r \lt |y-x_{0}|\text{,}\) it follows from
\begin{equation*} \vert a_{n}( z-x_{0})^{n}\vert \le \vert a_{n} (y-x_{0})^{n} \vert \frac{\vert z-x_{0}\vert^{n}}{\vert y-x_{0}\vert^{n}} \le C \left(\frac{r}{\vert y-x_{0}\vert}\right)^{n} \end{equation*}
for all \(z\) with \(|z-x_{0}| \le r\) and the comparison test with the geometric series \(\sum_{n=0}^{\infty} \left(\frac{r}{\vert y-x_{0}\vert}\right)^{n}\) that \(\sum_{n=0}^{\infty} a_{n}(z-x_{0})^{n}\) converges absolutely and uniformly for \(z\) such that \(|z-x_{0}|\le r\text{.}\) Since we can take any such \(r \lt |y-x_{0}|\text{,}\) we conclude that \(\sum_{n=0}^{\infty} a_{n}(z-x_{0})^{n}\) converges absolutely at any \(z\) with \(|z-x_{0}| \lt |y-x_{0}|\text{.}\)
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Remark 3.1.3.

The proof above shows that the Proposition holds for any complex \(z\) such that \(|z-x_{0}|\le r \lt |y-x_{0}|\) so the domain of convergence is a round disc centered at \(x_{0}\text{.}\) The the disc \(D(x_{0}, R)\) with radius \(R\) is the largest disc centered at \(x_{0}\) in which the power series converges.
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Definition 3.1.4. Radius of Convergence of a Power Series.

For any power series \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\text{,}\) we define
\begin{equation*} R =\sup\{ |x-x_{0}|: \sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n} \text{ converges}\} \end{equation*}
as the radius of convergence of this power series.
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Remark 3.1.5.

Note that \(R\) could be \(0\) or \(\infty\text{,}\) and that, if \(0 \lt R \lt \infty\text{,}\) then
\begin{align*} (a). \amp\text{ for any $0 \lt r \lt R$, the series absolutely and uniformly for $x$ such that $|x-x_{0}|\le r$,} \\ \amp \text{ thus $ \sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}$ defines a continuous function for $x$ such that $|x-x_{0}| \lt R$;}\\ (b). \amp\text{ for $z$ such that $|x-x_{0}|\gt R$, the series diverges at $x$.} \end{align*}
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Proof.

This follows from the root test: when
\begin{equation*} \vert x-x_{0}\vert \limsup_{n\to\infty} \vert a_{n}\vert^{1/n}= \limsup_{n\to\infty} \vert a_{n}(x-x_{0})^{n}\vert^{1/n} \lt 1, \end{equation*}
the power series \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\) converges absolutely, while when \(\limsup_{n\to\infty} \vert a_{n}(x-x_{0})^{n}\vert^{1/n} \gt 1\text{,}\) it diverges, for, this condition would imply the existence of a subsequence \(n_{k}\to \infty\) such that \(\vert a_{n_{k}}(x-x_{0})^{n_{k}}\vert^{1/n_{k}}\ge 1\text{,}\) which would then imply \(\vert a_{n_{k}}(x-x_{0})^{n_{k}}\vert \ge 1\text{,}\) so \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\) can’t converge at such an \(x\text{.}\)
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Whether \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\) converges for \(x\) with \(|x-x_{0}|=R\) depends on the particular series, as shown by the following examples
\begin{equation*} \sum_{n=0}^{\infty} x^{n},\quad \sum_{n=1}^{\infty}\frac{x^{n}}{n},\quad \sum_{n=1}^{\infty}\frac{x^{n}}{n^{2}}. \end{equation*}
All three have \(1\) as their radius of convergence; but the first series diverges at \(x=\pm 1\text{,}\) the second series converges (conditionally) at \(x=-1\) but diverges at \(x=1\text{,}\) and the third series converges absolutely for all \(x\) with \(|x|=1\text{.}\) Moreover, even though \(\sum_{n=0}^{\infty} x^{n}\) does not converge at \(x=-1\text{,}\) as \(x\to -1+0\text{,}\) \(\sum_{n=0}^{\infty} x^{n}=(1-x)^{-1}\to \frac 12\text{.}\) What about the limiting behavior of \(\sum_{n=1}^{\infty}\frac{x^{n}}{n}\) as \(x\to -1+0\text{?}\) This is answered by the Abel’s Theorem to be discussed in the next section.
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Since \(\limsup_{n\to\infty} \vert a_{n}\vert^{1/n}\) may not always be easy to evaluate, one often needs to provide an estimate for it.
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Exercise 3.1.7.

  1. Prove that \(\limsup_{n\to\infty} \sqrt[n]{|a_n|} \le \limsup_{n\to \infty} |\frac{a_{n+1}}{a_n}|\text{.}\) (One may assume that \(a_n \ne 0\) for all \(n\text{;}\) for, otherwise, the right hand side is \(\infty\text{.}\))
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  2. Prove that \(\liminf_{n\to \infty} |\frac{a_{n+1}}{a_n}| \le \liminf_{n\to\infty} \sqrt[n]{|a_n|}\text{.}\)
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  3. If \(\lim_{n\to \infty} |\frac{a_{n+1}}{a_n}|\) exists, then \(\lim_{n\to\infty} \sqrt[n]{|a_n|}\) exists, and equals \(\lim_{n\to \infty} |\frac{a_{n+1}}{a_n}|\text{.}\) (The converse is not true; construct some examples.)
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Remark 3.1.9.

In computing the radius of convergence of a power series, one often has to make some adaptation. For instance, to determine the radius of convergence of
\begin{equation*} \sum_{m=0}^{\infty} (-1)^{m}\frac{x^{2m+1}}{(2m+1)!}, \end{equation*}
if one applies the radius of convergence formula directly, then one needs to identify
\begin{equation*} a_{n}=\begin{cases} \frac{(-1)^{m}}{(2m+1)!} \amp \quad\text{ if } n=2m+1 \\ 0 \amp \quad\text{ if } n \text{ is even}, \end{cases} \end{equation*}
and evaluate \(\limsup_{n \to\infty} \sqrt[n]{|a_{n}|}\text{,}\) which would involve evaluation of
\begin{equation*} \limsup_{m \to\infty} \{(2m+1)!\}^{\frac{1}{2m+1}}\text{,} \end{equation*}
which can be done but requires some work. One could apply the ratio test but needs some adaptation, as \(a_{n}=0\) for all even \(n\)’s. One could treat \(x^{2m+1}\) as the \(m\)-th term, in stead of \((2m+1)\)-th term. Then one only needs to make sure that
\begin{equation*} |x|^{2} \limsup_{m\to \infty} \frac{(2m+1)!}{(2m+3)!} < 1. \end{equation*}
This turns out to hold for any \(x\text{.}\) Thus the radius of convergence here is \(\infty\text{.}\)
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Exercise 3.1.10. Radius of Convergence of a Power Series.

Suppose that the radius of convergence of \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n}\) is \(R\text{.}\) What is the radius of convergence of the series \(\sum_{n=0}^{\infty} a_{n}(x-x_{0})^{2n}\) and \(\sum_{n=0}^{\infty} a_{n}^{2}(x-x_{0})^{n}\) respectively?
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