\documentclass[a4paper,12pt]{article}
\begin{document}

\noindent {\bf Question}

\noindent Give five different examples of sequences that are
bounded but not convergent.
\medskip

\noindent {\bf Answer}
\begin{enumerate}
\item $\{ a_n =(-1)^n\}$, bounded above by $1$ and bounded below by
$-1$, hence bounded.  This sequence fails the Cauchy criterion,
since $|a_n -a_{n+1}| =2$ for all $n$, and so diverges.
\item $\{ \sin(n)\}$, bounded above by $1$ and bounded below by $-1$,
hence bounded.  Though it seems fairly clear why this sequence
diverges, the actual proof is a bit subtle, and we do not give it
here. 
\item $\{ 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, \ldots\}$, bounded
above by $1$ and bounded below by $0$, hence bounded.  Arbitrarily
far out in the sequence, there are consectutive terms taking the
values $0$ and $1$, so the sequence fails the Cauchy criterion and
hence diverges.
\item $\{ a_n = \mbox{ the ${\rm n}^{\rm th}$ digit of $\pi$} \}$,
bounded above by $9$ and bounded below by $0$, hence bounded. Does
not converge, because the only way for a sequence of integers to
converge is for it to be {\bf eventually constant}, that is,
constant past some index, which in this case would then imply that
$\pi$ is a repeating decimal, hence a rational number, which it
isn't.  (In fact, fixing an irrational number $x$ and taking $a_n$
to be the $n^{th}$ digit of the decimal expansion of $x$ gives a
sequence that is bounded but not convergent, by the same
argument.)
\item $\{ a_n = \mbox{ the ${\rm n}^{\rm th}$ digit of the rational
number $\frac{1}{7} = .\overline{142857}$} \}$, using the same
argument as above (which works for rational numbers, as long as
the length of the repeating section in the decimal expansion is
longer than one digit).
\end{enumerate}

\end{document}
