From 24f8c733a6082ad815009d53f418d44b382a2dfc Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Sat, 12 Nov 2022 14:43:58 -0500
Subject: 8 homeworks from analysis

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+\font\bigbf=cmbx12 at 24pt
+\newfam\bbold
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\def\bb#1{{\fam\bbold #1}}
+\input color
+
+\newcount\pno
+\def\tf{\advance\pno by 1\smallskip\bgroup\item{(\number\pno)}}
+\def\endtf{\egroup\medskip}
+\def\absolute{\centerline{{\color{blue} Absolute}\qquad Conditional\qquad Divergent}}
+\def\conditional{\centerline{Absolute\qquad{\color{blue} Conditional}\qquad Divergent}}
+\def\divergent{\centerline{Absolute\qquad Conditional\qquad{\color{blue} Divergent}}}
+
+\newcount\qno
+\long\def\question#1{\ifnum\qno=0\else\vfil\eject\fi\bigskip\pno=0\advance\qno by 1\noindent{\bf Question \number\qno.} {\it #1}}
+
+\def\intro#1{\medskip\noindent{\it #1.}\quad}
+\def\proof{\intro{Proof}}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+
+\let\bu\bullet
+
+{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad
+HOMEWORK \#5}}
+
+\question{Are the series absolutely convergent, conditionally
+convergent, or divergent? No proof required. (Cauchy Criteria can help
+on some of these.)}
+
+\tf
+$$\sum_n {\sin n\over n^2}$$
+\endtf
+
+\absolute
+
+\tf
+$$\sum_n {(-1)^n\over n\log n}$$
+\endtf
+
+\conditional
+
+\tf
+$${3\over 4}-{4\over 6}+{5\over 8}-{6\over 10}+{7\over 12}-\cdots$$
+\endtf
+
+\divergent
+
+\tf
+$$\sum_n {(-1)^n\over n(\log n)^2}$$
+\endtf
+
+\conditional
+
+\question{Show that the series $$\sum_n {(-1)^{n+1}\over n} = 1-{1\over
+2}+{1\over 3}-{1\over 4}+\cdots$$ is convergent. (This is the simplest
+instance of the Alternating Series Test (Thm 2.7.7). But you can't just
+cite that, since the Theorem is not proved in the text. You can follow
+any of the proof strategies outlined in Exercise 2.7.1. Or come up with
+your own, tailored to this particular example.}
+
+\proof
+We start by stating $$s_n = 1-{1\over 2}+{1\over 3}-{1\over 4}+\cdots\pm
+{1\over n}.$$
+Next, we calculate, for $n\geq 2k,$ $$s_n-s_{2k} = {1\over 2k+1} -
+{1\over 2k+2} +\cdots \pm {1\over n} = {1\over (2k+1)(2k+2)} + \cdots
+\pm {1\over n} > -{1\over n} > -{1\over 2k}.$$
+By similar reasoning, $s_n-s_{2k+1}<{1\over 2k}.$
+
+Therefore, $s_{2k}-{1\over 2k} < s_n < s_{2k+1}+{1\over 2k},$
+and since $|s_{2k+1}+{1\over 2k}-s_{2k}+{1\over 2k}| \leq {3\over 2k},$
+we can show for both that
+$$|s_n-s_{2k}| \leq {3\over 2k}\qquad |s_n-s_{2k+1}| \leq
+{3\over 2k}.$$
+This means the sequence is Cauchy and thus convergent.
+
+\endproof
+
+\question{(2.7.9, Ratio Test) Given a series $\sum_n a_n,$ with $a_n
+\neq 0,$ the Ratio Test states that if
+$$\limsup_n \left|{a_n\over a_{n+1}}\right| = r < 1,$$
+then the series $$\sum_n a_n$$ converges absolutely. Take these steps to
+verify this statement.}
+
+\item{a.}  Show that for $r<s<1,$ there is $N$ so that for all $n\geq
+N,$ we have
+$$\left| {a_n\over a_{n+1}} \right| < s.$$
+
+\proof
+The definition of $\limsup$ tells us that for some $N,$ that for all
+$n\geq N,$
+$$\left| {a_{n+1}\over a_n} \right| \leq r < s.$$
+\endproof
+
+\item{b.} Conclude that for $n > N,$ that $|a_n| < |a_N|s^{n-N}.$
+
+\proof
+We will show this by induction.
+
+Note that $$\left|{a_{n+1}\over a_n}\right| = {|a_{n+1}|\over |a_n|}.$$
+(This can be shown by a sort of casework: the signs of $a_n$ and
+$a_{n+1}$ can be independently flipped without affecting the expression,
+and the positive case is trivial)
+
+Using the same value of $N$ as in the previous question, we show the
+base case for $n = N+1,$
+$${|a_{N+1}|\over |a_N|} < s \Longrightarrow |a_n| < s|a_N|
+\Longrightarrow |a_n| < |a_N|s^{n-N}.$$
+For the inductive step, we assume for $n > N,$ that $|a_n| <
+s^{n-N}|a_N|,$ and then use ${|a_{n+1}|\over |a_n|} < s$ to show $|a_{n+1}|
+< s|a_n| < s^{n-N+1}|a_N|,$ completing the inductive step.
+\endproof
+
+\item{c.} Conclude that $\sum_n a_n$ converges absolutely.
+
+\proof
+Taking $N$ again as in previous questions,
+$$\sum_{n>N} |a_n| < \sum_{n>N} |a_N|s^{n-N} = {|a_N|\over 1-s},$$ and
+since the sequence is positive, the partial series are monotone
+increasing, giving convergence by the Monotone Convergence Theorem.
+
+The geometric sum is from the book.
+\endproof
+
+\bye
-- 
cgit 

