\magnification=\magstep1
\centerline{Mathematics 242--Principles of Analysis}
\centerline{Solutions for Practice Questions for Exam 2}
\centerline{October 31, 2005}
\bigskip
\noindent
I.  
\item{A)}  This is impossible by the Order Limit Theorem (2.3.4).
If $a_n < 0$ for all $n$ and $(a_n)$ converges to $b$ then
$b \le 0$.  
\item{B)}  Example: $x_n = (-1)^n$.  Note $x_n^2 = 1$ for all $n$.
\item{C)}  Example: $x_n = n$, $y_n = {1\over n^3}$.  Then
$\sum_n x_ny_n = \sum_n {n\over n^3} = \sum_n {1\over n^2}$ which 
converges.
\item{D)}  Example:  $x_n = n$ (or any other monotone increasing
sequence that is not bounded above).
\item{E)}  This is impossible by the result of Theorem 2.7.6 in 
the text.  
\item{F)}  Example: $\sum_{n=1}^\infty (-1)^n$.  The partial sums are 
$-1,0,-1,0,\cdots$.
\item{G)}  This is impossible because if you have any single
convergent subsequence, then any subsequence {\it of that 
subsequence} is also convergent, and to the same limit as
the first subsequence (Theorem 2.5.2 in the text).
\bigskip
\noindent
II.
\item{A)}  This is a geometric series with first term $-8/27$ 
and ratio $-2/3$.  The sum is 
${-8/27\over 1 + 2/3} = {-8\over 45}$
\item{B)}  For all $n \ge 2$, 
$${1\over n(n-1)} = {-1 \over n} + {1 \over n-1}$$
So for $m \ge 2$ the $m$th partial sum is 
$$\eqalign{
s_m &= {1\over 2 \cdot 1} + {1\over 3\cdot 2} + \cdots + {1\over m(m-1)}\cr
&= {1\over 1} - {1\over 2} + {1\over 2} - {1\over 3} + \cdots +
{1\over m-1} - {1\over m}\cr
&= 1 - {1\over m}.\cr}$$
(This is called a ``telescoping'' series because of these
cancellations.)  Then 
$$\sum_{n=2}^\infty {1\over n(n-1)} = 
\lim_{m\to\infty}\left(1 - {1\over m}\right) = 1.$$

\noindent
III.  
\item{A)}  $\sum_{k=1}^\infty {1\over k^{2/3}}$ is divergent
by comparison with the harmonic series:  ${1\over k^{2/3}} > {1\over k}$
for all $k \ge 1$.  
For all $k\in {\bf N}$, we have $k^{2/3} < (k+1)^{2/3}$, so
${1\over k^{2/3}} > {1\over (k+1)^{2/3}}$.  Moreover 
$\lim_{k\to\infty} {1\over k^{2/3}} = 0$.  Hence by the Alternating
Series Test, $\sum_{k=1}^\infty {(-1)^{k+1}\over k^{2/3}}$ converges.
The series is {\it conditionally convergent}.
\item{B)}  For the Ratio Test, we look at 
$$\eqalign{
\lim_{n\to\infty} \left|{a_{n+1}\over a_n}\right| 
&= \lim_{n\to\infty} \left|{(-1)^{n+1}9^{n+1}\over (2n+2)!} \cdot 
{(2n)!\over (-1)^n 9^n}\right|\cr
&=\lim_{n\to\infty} {9\over (2n+2)(2n+1)}\cr
&=0\cr}$$
Since this is $< 1$, the Ratio Test implies the series
is {\it absolutely convergent}.
\item{C)}  This series is {\it divergent} because
$\lim_{k\to\infty} {k^2 - 1\over k^2 + 2k + 4} = 1\ne 0$
(see Theorem 2.7.3 in text).
\item{D)}  For all $k \in {\bf N}$, $k < 2^k$.  Hence
${k\over 5^k} < \left({2\over 5}\right)^k$.  Since
$\sum_k \left({2\over 5}\right)^k$ is a geometric series
with ratio $0 < r = 2/5 < 1$, it is convergent.  So 
the given series converges too by the Comparison Test.
\item{E)}  Diverges by comparison with $\sum_k {1\over 2\sqrt{k+1}}$.
\bigskip
\noindent
IV. 
\item{A)}  If $(x_n)$ is convergent with limit $a$, 
then for all $\varepsilon > 0$, 
there exist $N \in {\bf N}$ such that $|x_n - a| < \varepsilon/2$
whenever $n \ge N$.  But then for that $N$, 
$$|x_n - x_{n+1}| = |(x_n - a) - (x_{n+1} - a)| \le |x_n - a| + |x_{n+1} - a|
< \varepsilon/2 + \varepsilon/2 = \varepsilon.$$
Hence $(x_n)$ is pC.
\item{B)} The converse of the statement in A is {\it false}.
Here is a counterexample.  Consider the sequence $(s_n)$ of partial
sums of the {\it divergent} harmonic series $\sum_k {1\over k}$.
$(s_n)$ is a divergent sequence since $\lim_{n\to\infty} s_n = +\infty$.
But $s_{n+1} - s_n = {1\over n+1}$.  So given any $\varepsilon > 0$,
there exists $N\in {\bf N}$ such that 
$|s_{n+1} - s_n| < \varepsilon$ whenver $n\ge N$ -- just
take $N \ge {1\over \varepsilon}$, then $n \ge N$
implies $|s_{n+1} - s_n| = {1\over n+1} < {1\over n} \le {1\over N} < \varepsilon$.
\bigskip
\noindent
V.  
\item{A)}  Since $\lim_{n\to\infty} na_n = \ell \ne 0$, and $a_n > 0$, the Order
Limit Theorem implies that $\ell > 0$.   For all
$\varepsilon > 0$, there exists $N\in {\bf N}$ such that 
$|na_n - \ell| < \varepsilon$ for all $n \ge N$.  Let $\varepsilon$
be small enough that $\ell - \varepsilon > 0$.  Then
$n a_n > \ell - \varepsilon$ whenever $n \ge N$, so 
$a_n > {\ell - \varepsilon\over n}$ whenever $n \ge N$.  This implies 
that $\sum_{n=N}^\infty a_n$ diverges, by comparison with the 
series $\sum_{n=N}^\infty (\ell - \varepsilon){1 \over n}$, which is a
constant times the ``tail'' of the harmonic series.
As a consequence, the whole series $\sum_{n=1}^\infty a_n$ diverges
too.
\item{B)}  Since $\lim_{n\to\infty} n^2 a_n = \ell \ne 0$, 
and $a_n > 0$, the Order
Limit Theorem implies that $\ell > 0$.   For all
$\varepsilon > 0$, there exists $N\in {\bf N}$ such that 
$|n^2 a_n - \ell| < \varepsilon$ for all $n \ge N$.  
Then $n^2 a_n < \ell + \varepsilon$ whenever $n \ge N$, so 
$a_n < {\ell - \varepsilon\over n^2}$ whenever $n \ge N$.  This implies 
that $\sum_{n=N}^\infty a_n$ converges, by comparison with the 
series $\sum_{n=N}^\infty (\ell + \varepsilon){1 \over n^2}$, which is a
constant times the ``tail'' of the series $\sum_n {1\over n^2}$.
As a consequence, the whole series $\sum_{n=1}^\infty a_n$ converges
too.
\vfill\eject
\noindent
VI.
\item{A)}  $S$ is nonempty by the Bolzano-Weierstrass theorem.
If $-M \le x_n \le M$ for all $n$ and $s\in S$, 
then $s = \lim_k x_{n_k}$, for some subsequence $x_{n_k}$.  
The Order Limit Theorem implies that because
$-M \le x_{n_k} \le M$,  $-M \le s\le M$ too.
Hence $S$ is bounded above and below.
\item{B)}  $\lim\sup\ (-1)^n + {1\over n} = 1$  (look at the 
even-numbered terms).  The odd-numbered terms show that $-1\in S$
for this sequence too. 
\item{C)}  (by contradiction)  Suppose that there exists
some $\varepsilon > 0$ such that infinitely many terms of the
sequence satsify $x_n > t + \varepsilon$.  From these
infinitely many terms, we select a subsequence $x_{n_k}$
with $x_{n_k} > t + \varepsilon$ for all $k\in {\bf N}$.
The $(x_{n_k})$ is also a bounded sequence, so the Bolzano-Weierstrass
Theorem implies that it has a convergent subsequence $x_{n_{k_\ell}}$. 
Say that convergent subsequence has limit $s$. Then $s\in S$ 
since $x_{n_{k_\ell}}$ is also a subsequence of the original $x_n$.
The Order Limit Theorem implies $s \ge t + \varepsilon > t$.  This 
contradicts the definition of $t$ as $t = \sup S$.  
\bye
