8 homeworks from analysisHEADmaster

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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\false{\centerline{TRUE\qquad{\color{blue} FALSE}}}
+\def\true{\centerline{{\color{blue} TRUE}\qquad FALSE}}
+
+\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 \#4, Part I}}
+
+\question{Answer True or False. No justification is needed.}
+
+\tf
+If $\lim a_n = 0$ and $\{b_n\}$ is a bounded sequence, then $a_nb_n$ is
+a convergent sequence.
+\endtf
+
+\true
+
+\tf
+If $\limsup_n a_n = 1$ and $\rho>1,$ then there is an $N>1$ such that
+$a_n<\rho$ for all $n>N.$
+\endtf
+
+\true
+
+\tf
+If $(a_n)$ is a Cauchy sequence, then $|a_n|$ is a Cauchy sequence.
+\endtf
+
+\true
+
+\tf
+If $(a_n)$ is a Cauchy sequence, then $\lfloor a_n\rfloor$ is a Cauchy
+sequence.
+\endtf
+
+\false
+
+\tf
+If $(a_n)$ is a Cauchy sequence, then $\arctan(a_n)$ is a Cauchy
+sequence.
+\endtf
+
+\true
+
+\question{(Exercise 2.6.5) Consider the following (invented) definition:
+a sequence $(s_n)$ is pseudo-Cauchy if, for all $\epsilon>0,$ there
+exists an $N$ such that if $n>N,$ then $|s_n-s_{n+1}|<\epsilon.$ Supply
+a proof for the valid statements and a counterexample for the false
+statements.}
+
+\item{a.} Pseudo-Cauchy sequences are bounded.
+
+\intro{Disproof}
+False. $s_n = \log n$ is clearly unbounded, but for all $\epsilon > 0,$
+we can choose some $N > 1/\epsilon.$
+
+For all $n > N > 1/\epsilon,$ (or $\epsilon > 1/n$), starting from a
+result of the definition of the exponential,
+$$e^{1/n} > 1+1/n = (n+1)/n \Longrightarrow
+\epsilon > 1/n > \log(n+1) - \log(n) = |s_n-s_{n+1}|.$$
+
+\endproof
+
+\item{b.} Bounded Pseudo-Cauchy sequences are convergent.
+
+\intro{Disproof}
+Consider $s_n = \log n,$ but modulated such that it ``bounces off'' the
+bounds of the interval $(0,1),$ or more directly,
+$s_n = |\log n\bmod 2-1|.$
+\endproof
+
+\item{c.} If $(x_n)$ and $(y_n)$ are pseudo-Cauchy, then $(x_n+y_n)$ is
+Pseudo-Cauchy as well.
+
+\proof
+Let $\epsilon > 0.$ We must have $N$ and $M$ such that for all $n >
+\max\{N,M\},$ for both sequences, $|x_n-x_{n+1}|<\epsilon/2$ and
+$|y_n-y_{n+1}|<\epsilon/2.$
+By the triangle inequality,
+$$|x_n+y_n-x_{n+1}-y_{n+1}|<|x_n-x_{n+1}|+|y_n-y_{n+1}|<\epsilon.$$
+\endproof
+
+\question{Give an example of each or explain why the request is
+impossible referencing the proper theorem(s).}
+
+\item{a.} Two series $\sum x_n$ and $\sum y_n$ that both diverge but
+where $\sum x_ny_n$ converges.
+
+\intro{Example.}
+Let $x_n = 1/n$ and $y_n = -1.$
+\endproof
+
+\item{b.} A convergent series $\sum x_n$ and a bounded sequence $(y_n)$
+such that $\sum x_n y_n$ diverges.
+
+\intro{Example.}
+Let $x_n = (-1)^n 1/n$ and $y_n = (-1)^n.$
+\endproof
+
+\item{c.} Two sequences $(x_n)$ and $(y_n)$ where $x_n$ and $(x_n+y_n)$
+both converge but $y_n$ diverges.
+
+\intro{Disproof.}
+If $x_n$ and $(x_n+y_n)$ converge, their difference should also
+converge, their difference being $x_n+y_n-x_n = y_n.$
+This is implied from the Algebraic Limit Theorem for Series.
+\endproof
+
+\item{d.} A sequence $(x_n)$ satisfying $0<x_n\leq 1/n$ where $\sum_n
+(-1)^n x_n$ diverges.
+
+\intro{Disproof.}
+This converges by the Alternating Series Test.
+\endproof
+
+\bye