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\title{Supplementary Notes:\\The Weierstrass $M$-test and Power Series%
\footnote{
\copyright 2019 L. Tu, Z. Nitecki and Tufts University
}
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%\author{Loring Tu and Zbigniew Nitecki}%
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%\copyright 2019 L. Tu, Z. Nitecki and Tufts University
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%\copyright 2019 L. Tu, Z. Nitecki and Tufts University
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%\begin{center}
%Math 135 \hfill Real Analysis I \hfill Fall 2019 \\
%\phantom{Prof.~L.~Tu}\hfill \large\textbf{The Weierstrass $M$-test and Power Series}%
%\footnote{
%\copyright 2019 L. Tu, Z. Nitecki and Tufts University
%}%
% \hfil\ \phantom{Prof.~L.~Tu}
%\end{center}

\bigskip
Recall that the \textbf{Taylor series} at $x=a$ of a function \fofx{} is the series of polynomials
\begin{multline*}
	\sum_{k=\textcolor{red}{0}}^{\infty}\frac{\fpnof{k}{a}}{k!}(x-a)^{k}=\\
	\fof{a}+\fpof{a}(x-a)+\frac{\fppof{a}}{2!}(x-a)^{2}+\cdots+\frac{\fpnof{n}{a}}{n!}(x-a)^{n}+\cdots
\end{multline*}
where \fpnof{k}{a} is the \kth{} derivative of \fofx{}%
\footnote{\fpnof{0}{a} is the ``undifferentiated ''
function \fof{a} and $k!$ is the factorial of $k$, with $0!=1$}
 at $x=a$.
 The partial sums of the Taylor series are the \textbf{Taylor polynomials} of \fofx{} at $x=a$.
 
 In this note, we study the convergence of series of this form:
 \begin{definition}\label{dfn:powers}
 	A \textbf{power series} is a series of polynomials of the form%
	\footnote{
	We have rendered the starting index $k=\textcolor{red}{0}$ to underline that a power series 
	can have a ``constant'' term, and it is convenient to have the index run over the non-negative integers, 
	despite Fitzpatrick's convention of always using the natural numbers to index a sequence.
	}
	\begin{equation*}
		\sum_{k=\textcolor{red}{0}}^{\infty}\cs{k}(x-a)^{k}.
	\end{equation*}
	The series is \textbf{centered at $x=a$}.
 \end{definition}
	We will for the most part focus on series centered at $x=0$
	\begin{equation*}
		\sum_{k=\textcolor{red}{0}}^{\infty}\cs{k}x^{k};
	\end{equation*}
	at the end we will use the substitution $x\mapsto(x-a)$ 
	to obtain related results for power series centered elsewhere.
	
	Our goal is to prove and explain the following picture:
	\begin{itemize}
		\item The set of points where the series $\sum_{k=\textcolor{red}{0}}^{\infty}\cs{k}x^{k}$ 
		converges is an interval $I$, called the \textbf{interval of convergence} of the series;
		\item $I$ consists of the singleton $\single{0}=\clint{0}{0}$, the whole line \opint{-\infty}{\infty}, 
		or an interval (open, closed, or half-open)
		whose endpoints are $\pm R$, where $0<R<\infty$; we call $R$ ($0\leq R\leq\infty$) the  
		\textbf{radius of convergence} of the series 
		\item the convergence of the series is absolute at any point interior to the interval of convergence,
		and uniform on any sequentially compact interval
		contained in $I$.
	\end{itemize}


\section{The Weierstrass $M$-test}\label{sec:MTest}
The following analogue of the Comparison Test for numerical series is a very useful tool for proving 
the uniform convergence of a series of functions.

\begin{theorem}[Weierstrass M-test] Suppose a sequence of functions \Rfunct{\fs{k}}{D}, $k=0,1, \dots,$
satisfies the estimates
\begin{equation*}
	\abs{\fsof{k}{x}}\leq\Ms{k},\ k=0,1,\dots,\text{ for all }x\in D
\end{equation*}
where the constants $\Ms{k}>0$ satisfy
\begin{equation*}
	\infsum{k}{0}{\Ms{k}}<\infty.
\end{equation*}
Then the series
\begin{equation*}
	\infsum{k}{0}{\fsof{k}{x}}
\end{equation*}
converges uniformly on $D$.
\end{theorem}

\begin{proof}
	Since the sum \infsum{k}{0}{\Ms{k}} converges, its partial sums $\Ss{N}=\finsum{k}{0}{N}{\Ms{k}}$
	form a Cauchy sequence:
	\begin{equation*}
		\forall\epsgo,\exists N\in\Nat \suchthat m,n\geq N\imply \abs{\Ss{m}-\Ss{n}}<\eps.
	\end{equation*}
	For $N\leq n<m$, $\abs{\Ss{m}-\Ss{n}}=\finsum{k}{n+1}{m}{\Ms{k}}$;  
	the partial sums of the series \infsum{k}{0}{\fsof{k}{x}} satisfy, for $N\leq n<m$,
	\begin{equation*}
		\abs{\finsum{k}{0}{m}{\fsof{k}{x}}-\finsum{k}{0}{n}{\fsof{k}{x}}}=\abs{\finsum{k}{n+1}{m}{\fsof{k}{x}}}
		\leq\finsum{k}{n+1}{m}{\abs{\fsof{n}{x}}}\leq\finsum{k}{n+1}{m}{\Ms{k}}<\eps,
	\end{equation*}
	showing that the sequence of partial sums of \infsum{k}{0}{\fsof{k}{x}} is uniformly Cauchy on $D$,
	and hence the series is uniformly convergent on $D$.
\end{proof}

\section{Power Series}\label{sec:PowerS}
\subsection{Interval of Convergence}
\begin{theorem}\label{thm:int}
	The power series \infsum{k}{\textcolor{red}{0}}{\cs{k}x^{k}} always converges at $x=0$; 
	if it converges at $x=b\neq0$, then the series converges 
	absolutely at any point $x$ in the open interval \opint{-\abs{b}}{\abs{b}};
	furthermore, for any 
	$r$ such that $0< r<\abs{b}$, the series 
	converges uniformly on the closed interval \clint{-r}{r}.
\end{theorem}

\begin{proof}
	That the series converges at $x=0$ follows from the fact that every term beyond $\cs{0}x^{0}=\cs{0}$
	is zero.
	
	If \infsum{k}{0}{\cs{k}b^{k}} converges, then by the Divergence Test the sequence \single{\cs{k}b^{k}}
	converges (to $0$) and hence is bounded: let $C$ be an upper bound for the absolute values of these
	terms:
	\begin{equation*}
		\abs{\cs{k}b^{k}}\leq C\text{ for all }k=0,1,\cdots.
	\end{equation*}
	
	Suppose $\abs{x}\leq r<\abs{b}$;  then
	\begin{equation*}
		\abs{\cs{k}x^{k}}
		=\abs{(\cs{k}b^{k})\left(\frac{x}{b}\right)^{k}}
		\leq\abs{\cs{k}b^{k}}\left(\frac{r}{\abs{b}}\right)^{k}
		\leq C\left(\frac{r}{\abs{b}}\right)^{k}.
	\end{equation*}
	Setting $\Ms{k}=C\left(\frac{r}{\abs{b}}\right)^{k}$ 
	and noting that the series \infsum{k}{0}{C\left(\frac{r}{\abs{b}}\right)^{k}} is a geometric series 
	with ratio less than $1$, it follows
	from the Weierstrass M-Test that \infsum{k}{\textcolor{red}{0}}{\cs{k}x^{k}} converges absolutely and 
	uniformly on \clint{-r}{r}.
\end{proof}

\begin{remark}\label{rmk:interval}
	An interval can be characterized as a set $S\subset\Reals$ with the property that if
	$x,y\in S$ and $x<z<y$ then $z\in S$.
\end{remark}

Combining \refer{thm}{int} and \refer{rmk}{interval} we obtain

\begin{corollary}\label{cor:Interval}
	The convergence set of a power series centered at $x=0$
	\begin{equation*}
		I=\setbld{x\in\Reals}{ \infsum{k}{\textcolor{red}{0}}{\cs{k}x^{k}}\text{ converges}}
	\end{equation*}
	is an interval (open or closed or half-open). 
	Let $R=\sup I$. The possibilities are
	\begin{description}
		\item[$\boldsymbol{R=0}$] $I=\single{0}=\clint{0}{0}$: The series converges at $x=0$ and diverges 
		if $x\neq0$.
		
		\item[$\boldsymbol{0<R<\infty}$] $I$ is an interval with finite endpoints $\pm R$, The series converges
		absolutely at every point $x$ with $\abs{x}<r$ and diverges at every point $x$ with $\abs{x}>R$;
		convergence is uniform on \clint{-r}{r} for every $r<R$.
%		;  it diverges at every $x$ with $\abs{x}>R$.  

		Convergence at the two endpoints $x=\pm R$ is not determined: depending on the series, it
		can converge conditionally, converge absolutely, or diverge at $x=R$ and (independently) at $x=-R$.
		
		\item[$\boldsymbol{R=\infty}$] The series converges absolutely at every $x\in \Reals$, 
		uniformly on any bounded interval.		
	\end{description}
	The number $R=\sup I$ is called the \textbf{radius of convergence} of the series 
	 \infsum{k}{\textcolor{red}{0}}{\cs{k}x^{k}}.
\end{corollary}

%\begin{proof}
%	
%\end{proof}

\subsection{Finding the interval of convergence (OPTIONAL)}

An application of the Ratio Test can in many cases give us a way to determine the radius of convergence of a 
power series.

\begin{prop}\label{prop:FindR}
	Suppose the coefficients of the power series \infsum{k}{\textcolor{red}{0}}{\cs{k}x^{k}} satisfy
	\begin{equation*}
		\abs{\frac{\cs{k+1}}{\cs{k}}}\to\rho.
	\end{equation*}
	
	Then the radius of convergence of the series is $R=\recip{\rho}$, where we mean
	 $R=\recip{0}=\infty$ if $\rho=0$
	and $R=\recip{\infty}=0$ if $\rho=\infty$.
\end{prop}

\begin{proof}
	The series always converges at $x=0$, as noted earlier.
	
	If $0<\rho<\infty$, then for any $x\neq0$ we apply the Ratio Test to the numerical series
	\infsum{k}{\textcolor{red}{0}}{\cs{k}x^{k}}:
	\begin{equation*}
		\abs{\frac{\cs{k+1}x^{k+1}}{\cs{k}x^{k}}}=\abs{\frac{\cs{k+1}}{\cs{k}}}\abs{\frac{x^{k+1}}{x^{k}}}
		\to \rho\abs{x}=\abs{\frac{x}{R}}
	\end{equation*}
	By the ratio test, the series converges if $\abs{\frac{x}{R}}<1$ (\ie{} $\abs{x}<R$) 
	and diverges if $\abs{\frac{x}{R}}>1$ (\ie{} $\abs{x}>R$).
	
	If $\rho=0$, the ratio test tells us that the series converges for all $x\in\Reals$, since the ratio goes to $0<1$.	If $\rho=\infty$, for any $x\neq 0$ the terms of the series diverge to infinity. 
\end{proof}

%\begin{remark}\label{rmk:altTests}
	There are several more sophisticated tests that can be used to find the radius of convergence when 
	\refer{prop}{FindR} does not apply.  We state them without proof:
%	\begin{description}

		\textbf{Root Test:}  If $\sqrt[k]{\abs{\cs{k}}}\to\rho$ then the radius of convergence is $R=\recip{\rho}$
		as before.
		
		\textbf{limit superior:} If the ratios $\abs{\frac{\cs{k+1}}{\cs{k}}}$ do not converge, we can replace their
		limit with the following:
%		\textbf{limit superior}, defined as follows 
\begin{definition}\label{dfn:limsup}
		Suppose \single{\rs{k}} is a non-negative sequence.
		Given $K\in\Nat$, let 
		\begin{equation*}
			\ssub{K}=\sup\setbld{\rs{k}}{k\geq K}.
		\end{equation*}
		This is a decreasing sequence, since we are taking suprema over smaller sets.
		Since we assume $\rs{k}\geq 0$ for all $k$ the sequence  \single{\ssub{K}} is bounded 
		and monotone, hence converges.  Define the \textbf{limit superior} of \single{\rs{k}}as
		\begin{equation*}
			\limsup_{k} \rs{k}=\lim_{K}\ssub{K}.
		\end{equation*}
\end{definition}
		
		Then if we define $\rho$ to be the the limit superior instead of the limit of the ratios in the Ratio Test,
		we get a version which (by tweaking our arguments in the proof of \refer{prop}{FindR}) always 
		yields a value for the radius of convergence: if
		\begin{equation*}
			\rho=\limsup\abs{\frac{\cs{k+1}}{\cs{k}}},\ R=\recip{\rho}
		\end{equation*} 
		then the series always converges if $\abs{x}<R$ and diverges if $\abs{x}>R$.	
%	\end{description}
%\end{remark}

\subsection{General power series}

Our discussion has focused on power series centered at $x=0$, \infsum{k}{\textcolor{red}{0}}{\cs{k}x^{k}}.
To handle a power series centered at $x=a$ when $a\neq0$, \infsum{k}{\textcolor{red}{0}}{\cs{k}(x-a)^{k}},
we can make the substitution $y=x-a$ to obtain a new power series, \infsum{k}{\textcolor{red}{0}}{\cs{k}y^{k}},
centered at $y=0$.  We leave it to you to verify that applying \refer{cor}{Interval} to this new series leads 
to the following generalization to arbitrary power series:

\begin{theorem}\label{thm:Interval}
	The convergence set of a power series centered at $x=a$
	\begin{equation*}
		I=\setbld{x\in\Reals}{ \infsum{k}{\textcolor{red}{0}}{\cs{k}(x-a)^{k}}\text{ converges}}
	\end{equation*}
	is an interval (open or closed or half-open) whose midpoint is $a$. 
	
	The possibilities are
	\begin{description}
		\item[$\boldsymbol{R=0}$] $I=\single{a}=\clint{a}{a}$: The series converges at $x=a$ and diverges 
		if $x\neq a$.
		
		\item[$\boldsymbol{0<R<\infty}$] $I$ is an interval with finite endpoints $a\pm R$, The series converges
		absolutely at every point $x$ with $\abs{x-a}<r$ and diverges at every point $x$ with $\abs{x-a}>R$;
		convergence is uniform on \clint{a-r}{a+r} for every $r<R$.
%		;  it diverges at every $x$ with $\abs{x}>R$.  

		Convergence at the two endpoints, $x=a\pm R$, is not determined: depending on the series, it
		can converge conditionally, converge absolutely, or diverge at $x=a+R$ 
		and (independently) at $x=a-R$.
		
		\item[$\boldsymbol{R=\infty}$] The series converges absolutely at every $x\in \Reals$, 
		uniformly on any bounded interval.		
	\end{description}
\end{theorem}
	The number $R=\sup I$ is called the \textbf{radius of convergence} of the series 
	 \infsum{k}{\textcolor{red}{0}}{\cs{k}(x-a)^{k}}.  \refer{prop}{FindR} and its variants apply verbatim 
	 for finding this radius.








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