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{\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
