\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