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\noindent {\bf Question}

\noindent Give an example of a sequence that is bounded but not
convergent, or prove that no such sequence exists.  Also, give an
example of a sequence that is convergent but not bounded, or prove
that no such sequence exists.


\medskip

\noindent {\bf Answer}

\noindent The sequence $\{ a_n =(-a)^n\}$ is bounded below by $-1$
and bounded above by $1$, and so is bounded.  This sequence does
not converge, though; since $|a_n -a_{n+1}| =2$ for all $n$, this
sequence fails the Cauchy criterion, and hence diverges.

\medskip
\noindent For the other part, we know that every convergent
sequence is bounded.  This is the proposition below. (Note that
you are asked in this question to state and to write out the proof
of this proposition as set out in the note below.)

\medskip
\noindent {\bf Note:}

\noindent Proposition

\noindent Let $A =\{ a_n\}$ be a convergent sequence. Then, $A$ is
bounded.

\medskip
\noindent Proof

\noindent Set $a =\lim_{n\rightarrow\infty} a_n$, and apply the
definition of limit of a sequence with $\varepsilon =1$, so that
there exists $M >0$ so that $| a_n -a| <1$ for all $n >M$. In
particular, for $n >M$, we have that $a_n$ lies in the interval
$(a-1, a+1)$.  Let $s =\max( a_1,\ldots, a_M, a+1)$, and note that
$a_n\le s$ for all $n$.  In particular, $A =\{ a_n\}$ is bounded
above by $s$.

\medskip
\noindent Similarly, set $t =\min(a_1,\ldots, a_M, a-1)$, and note
that $t\le a_n$ for all $n$, so that $A =\{ a_n\}$ is bounded
below by $t$.

\medskip
\noindent Since $A$ is both bounded below and bounded above, it is
bounded. (Note that the choice of $\varepsilon =1$ is completely
arbitrary. Any positive number will work.)

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