Theorem 10.3.1.
Suppose \(\sum_{n=0}^\infty a_nc^n\) converges for some nonzero real number \(c\text{.}\) Then \(\sum_{n=0}^\infty a_nx^n\) converges absolutely for all \(x\) such that \(|x|\lt |c|\text{.}\)
Section 10.3 Radius of Convergence of a Power Series
We’ve developed enough machinery to look at the convergence of power series. The fundamental result is the following theorem due to Abel.
To prove Theorem 10.3.1 first note that by Problem 10.2.22, \(\limit{n}{\infty}{a_nc^n}=0\text{.}\) Thus \(\left(a_nc^n\right)\) is a bounded sequence. Let \(B\) be a bound: \(\abs{a_nc^n}\le B\text{.}\) Then
\begin{equation*}
\abs{a_nx^n}=\abs{a_nc^n\cdot\left(\frac{x}{c}\right)^n}\leq B\abs{\frac{x}{c}}^n\text{.}
\end{equation*}
We can now use the comparison test.
Problem 10.3.2.
Prove Theorem 10.3.1.
Corollary 10.3.3.
Suppose \(\sum_{n=0}^\infty a_nc^n\) diverges for some real number \(c\text{.}\) Then \(\sum_{n=0}^\infty a_nx^n\) diverges for all \(x\) such that \(|x|>|c|\text{.}\)
Problem 10.3.4.
Prove Corollary 10.3.3.
As a result of Theorem 10.3.1 and Corollary 10.3.3, we have the following: either \(\sum_{n=0}^\infty a_nx^n\) converges absolutely for all \(x\) or there exists some nonnegative real number \(r\) such that \(\sum_{n=0}^\infty a_nx^n\) converges absolutely when \(|x|\lt r\) and diverges when \(|x|>r\text{.}\) In the latter case, we call \(r\) the radius of convergence of the power series \(\sum_{n=0}^{\infty}a_{n} x^{n}\text{.}\) In the former case, we say that the radius of convergence of \(\sum_{n=0}^\infty a_nx^n\) is \(\infty\text{.}\) Though we can say that \(\sum_{n=0}^\infty a_nx^n\) converges absolutely when \(|x|\lt r\text{,}\) we cannot say that the convergence is uniform. However, we can come close. We can show that the convergence is uniform for \(|x|\leq b\lt r\text{.}\) To see this we will use the following result
Theorem 10.3.5.
The Weierstrass-\(M\) Test
Let \(\left(f_n\right)_{n=1}^\infty\) be a sequence of functions defined on \(S\subseteq\RR\) and suppose that \(\left(M_n\right)_{n=1}^\infty\) is a sequence of nonnegative real numbers such that
\begin{equation*}
\abs{f_n(x)}\leq M_n,\,\, \forall x\in S,\,\, n=1, 2, 3, \ldots\text{.}
\end{equation*}
If \(\sum_{n=1}^\infty M_n\) converges then \(\sum_{n=1}^\infty f_n(x)\) converges uniformly on \(S\) to some function (which we will denote by \(f(x)\)).
Sketch of Proof.
Since the crucial feature of the theorem is the function \(f(x)\) that our series converges to, our plan of attack is to first define \(f(x)\) and then show that our series, \(\sum_{n=1}^\infty f_n(x)\text{,}\) converges to it uniformly.
First observe that for any \(x\in S\text{,}\) \(\sum_{n=1}^\infty f_n(x)\) converges by the Comparison Test (in fact it converges absolutely) to some number we will denote by \(f(x)\text{.}\) This actually defines the function \(f(x)\) for all \(x\in S\text{.}\) It follows that \(\sum_{n=1}^\infty f_n(x)\) converges pointwise to \(f(x)\text{.}\)
Next, let \(\eps>0\) be given. Notice that since \(\sum_{n=1}^\infty M_n\) converges, say to \(M\text{,}\) then there is a real number, \(N\text{,}\) such that if \(n>N\text{,}\) then
\begin{equation*}
\sum_{k=n+1}^\infty M_k = \abs{\sum_{k=n+1}^\infty M_k} = \abs{M-\sum_{k=1}^n M_k}\lt \eps\text{.}
\end{equation*}
You should be able to use this to show that if \(n>N\text{,}\) then
\begin{equation*}
\abs{f(x) - \sum_{k=1}^n f_k(x)}\lt \eps, \, \, \forall x\in S\text{.}
\end{equation*}
Problem 10.3.6.
Use the ideas above to provide a formal proof of Theorem 10.3.5.
Problem 10.3.7.
(a)
Referring back to equation (1), show that the Fourier series
\begin{equation*}
\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\sin\left((2k+1)\pi x\right)
\end{equation*}
converges uniformly on \(\RR\text{.}\)
(b)
Does its differentiated series converge uniformly on \(\RR?\) Explain.
Problem 10.3.8.
Observe that for all \(x \in [-1,1]\) \(\abs{x}\le 1\text{.}\) Identify which of the following series converges pointwise and which converges uniformly on the interval \([-1,1]\text{.}\) In every case identify the limit function.
(a)
\(\displaystyle \sum_{n=1}^\infty\left(x^n-x^{n-1}\right)\)
(b)
\(\displaystyle \sum_{n=1}^\infty\frac{\left(x^n-x^{n-1}\right)}{n}\)
(c)
\(\sum_{n=1}^\infty\frac{\left(x^n-x^{n-1}\right)}{n^2}\)
Using the Weierstrass-\(M\) test, we can prove the following result.
Theorem 10.3.9.
Suppose \(\sum_{n=0}^\infty a_nx^n\) has radius of convergence \(r\) (where \(r\) could be \(\infty\) as well). Let \(b\) be any nonnegative real number with \(b\lt r\text{.}\) Then \(\sum_{n=0}^\infty a_nx^n\) converges uniformly on \([-b,b]\text{.}\)
Problem 10.3.10.
Prove Theorem 10.3.9.
Hint.
We know that \(\sum_{n=0}^\infty|a_nb^n|\) converges. This should be all set for the Weierstrass-\(M\) test.
To finish the story on differentiating and integrating power series, all we need to do is show that the power series, its integrated series, and its differentiated series all have the same radius of convergence. You might not realize it, but we already know that the integrated series has a radius of convergence at least as big as the radius of convergence of the original series. Specifically, suppose \(f(x)=\sum_{n=0}^\infty a_nx^n\)has a radius of convergence \(r\) and let \(|x|\lt r\text{.}\) We know that \(\,\sum_{n=0}^\infty a_nx^n\) converges uniformly on an interval containing \(0\)and \(x\text{,}\) and so by Corollary 10.2.4, \(\int_{t=0}^xf(t)\dx{ t}=\sum_{n=0}^\infty\left(\frac{a_n}{n+1}x^{n+1}\right)\text{.}\) In other words, the integrated series converges for any \(x\) with\(\,|x|\lt r\text{.}\) This says that the radius of convergence of the integated series must be at least \(r\text{.}\)
To show that the radii of convergence are the same, all we need to show is that the radius of convergence of the differentiated series is at least as big as \(r\) as well. Indeed, since the differentiated series of the integrated series is the original, then this would say that the original series and the integrated series have the same radii of convergence. Putting the differentiated series into the role of the original series, the original series is now the integrated series and so these would have the same radii of convergence as well. With this in mind, we want to show that if \(|x|\lt r\text{,}\) then \(\sum_{n=0}^\infty a_nnx^{n-1}\) converges. The strategy is to mimic what we did in Theorem 10.3.1, where we essentially compared our series with a converging geometric series. Only this time we need to start with the differentiated geometric series.
Problem 10.3.11.
Show that \(\sum_{n=1}^\infty nx^{n-1}\) converges for \(|x|\lt 1\text{.}\)
Hint.
We know that \(\sum_{k=0}^nx^k=\frac{x^{n+1}-1}{x-1}\text{.}\) Differentiate both sides and take the limit as \(n\) approaches infinity.
Theorem 10.3.12.
Suppose \(\sum_{n=0}^\infty a_nx^n\) has a radius of convergence \(r\) and let \(|x|\lt r\text{.}\) Then \(\sum_{n=1}^\infty a_nnx^{n-1}\) converges.
Problem 10.3.13.
Prove Theorem 10.3.12.
Hint.
Let \(b\) be a number with \(\abs{x}\lt b\lt r\) and consider \(\abs{a_nnx^{n-1}} =\abs{a_nb^n\cdot\frac{1}{b}\cdot
n\left(\frac{x}{b}\right)^{n-1}}\text{.}\) You should be able to use the Comparison Test and Problem 10.3.11.
