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diff --git a/lacey/hw5.tex b/lacey/hw5.tex new file mode 100644 index 0000000..08b8638 --- /dev/null +++ b/lacey/hw5.tex @@ -0,0 +1,130 @@ +\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 |