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\title{Math 3110: Homework 4}
%\author{Zach Norwood}
\date{Due Wednesday, 20 February 2019}

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\newtheorem {problem} {Problem}

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\begin{document}
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\noindent
You will be graded on both the accuracy and the clarity of your solutions.
One purpose of the homework is to give you an opportunity to practice your proof-writing skills.

You are welcome\,---\,encouraged, even!\,---\,to collaborate on homework, but you should not 
copy solutions from any source, nor should you submit anything that you don't understand.

\begin{problem}
 I'm thinking of ten different ten-digit numbers (e.g. $4860239777$).
 If you choose a number that's \emph{different} from all ten of my numbers, you win a prize.
 You can ask me ten yes-or-no questions about my numbers.
 Which ten questions do you ask, and how do you use the answers to make your choice?
 (\emph{Hint:} This is related to what we've proved in class recently.)
\end{problem}

\begin{problem}
For each item, provide an example (and prove that it works) or prove that no such example exists.
 \begin{enumerate} [(a)]
  \item A bounded sequence that does not converge to $\frac49$ but has a subsequence converging to $\frac49$.
  \item A monotone sequence that does not converge to $\frac49$ but has a subsequence converging
  to $\frac49$.
  \item A sequence with both an increasing subsequence and a decreasing subsequence that does not 
  converge.
  \item A bounded monotone sequence that does not converge.
  \item A sequence that does not converge and has no convergent subsequences.
  \item A bounded sequence with an unbounded subsequence.
 \end{enumerate}
\end{problem}

\begin{problem}
For each item, provide an example (and prove that it works) or prove that no such example exists.
 \begin{enumerate} [(a)]
  \item A sequence $(a_n)_{n\in\N}$ such that $3 < a_n < 4$ for all $n\in\N$ and $(a_n)_{n\in\N}$ has a
  subsequence converging to $3$ and another subsequence converging to $4$.
  \item 
     A sequence $(a_n)_{n\in\N}$ such that, for each $k\in\N\wo\{0\}$, there is a subsequence of $(a_n)_{n\in\N}$ 
   converging to $\tfrac1k$.
  \item
   A sequence $(a_n)_{n\in\N}$ such that, for each $k\in\N\wo\{0\}$, there is a subsequence of $(a_n)_{n\in\N}$ 
   converging to $\tfrac1k$, but there is no subsequence of $(a_n)_{n\in\N}$ converging to $0$.
  \item A sequence $(a_n)_{n\in\N}$ such that 
  for every real number $x$, the sequence $(a_n)_{n\in\N}$ has a subsequence that converges to $x$.
 \end{enumerate}
\end{problem}

\begin{problem}
 Suppose that $(a_n)_{n\in\N}$ is a sequence of nonnegative real numbers.
 Prove that the series $\displaystyle\sum_{n\ge0} a_n$ converges if and only if the series
 \[
  \sum_{n\ge0} 2^n a_{2^n} = a_1 + 2a_2 + 4a_4 + 8a_8 + \cdots
 \]
 converges.
 
 (\emph{Hint:} To prove the implication $\Leftarrow$, imitate the proof that the harmonic series diverges.)
\end{problem}

\begin{problem} [essentially Exercise 2.4.7 of the textbook]
 Let $(a_n)_{n\in\N}$ be a bounded sequence, and define the auxiliary \emph{limit superior}
 and \emph{limit inferior} sequences as follows:
 \[
   S_n = \sup\set{ a_k }{ k\ge n} \quad\text{and}\quad I_n = \inf\set{ a_k }{ k\ge n }.
 \]
 \begin{enumerate} [(a)]
  \item Show (in roughly one sentence, please) that $(S_n)_{n\in\N}$ is a decreasing sequence 
  and that $(I_n)_{n\in\N}$ is an increasing sequence.
 \item Using the assumption that $(a_n)_{n\in\N}$ is bounded, prove that $(S_n)_{n\in\N}$ converges.
 Convince yourself (but don't submit this with your homework) that $(I_n)_{n\in\N}$ also converges.
 \end{enumerate}
 We can therefore define the ``limsup'' and ``liminf'' of an arbitrary bounded sequence as follows:
 \begin{align*}
  \limsup a_n 	&= 	\lim_{n\to\infty} S_n = \lim_{n\to\infty} \left[ \sup_{k\ge n} a_k \right]\\
  \liminf a_n  	&= 	\lim_{n\to\infty} I_n = \lim_{n\to\infty} \left[ \inf_{k\ge n} a_k \right].
 \end{align*}
 \begin{enumerate} [(a), resume]
  \item Prove that $\liminf a_n \le \limsup a_n$ for every bounded sequence $(a_n)_{n\in\N}$,
  and give an example of a sequence for which the inequality is strict.
  \item Prove that $\liminf a_n = \limsup a_n$ if and only if $(a_n)_{n\in\N}$ converges.
  In this case, $\liminf a_n = \lim a_n = \limsup a_n$.
 \end{enumerate} 
\end{problem}



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