\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions -- Midterm Exam 2}
\centerline{November 1, 2004}
\bigskip
\noindent
I.  
\item{A)}  The set $A$ is the union of two open intervals:
$A = (-2,-1) \cup (1,2)$, so the intersection is
$S = A \cap B = (1,3/2)$.
\item{B)}  $\sup(S) = 3/2$ and $\inf(S) = 1$.  
\bigskip
\noindent
II.
\item{A)}  Completeness (Least Upper Bound) Axiom:  For every nonempty subset 
$S$ of ${\bf R}$ that is bounded above, there exists a least upper bound
$\sup(S) \in {\bf R}$.  (That is, if there exists $M \in {\bf R}$
such that $M \ge x$ for all $x \in S$, then there exists $s = \sup(S) \in {\bf R}$
with the properties that $s \ge x$ for all $x \in S$, and if $s' < s$, then
there exists some $x \in S$ such that $x > s'$.  {\it However}, it is
not necessary to say all of this!) 
\item{B)} Since $(x_n)$ is bounded above, the set $S = \{x_n : n \in {\bf N}\}$
is a bounded subset of ${\bf R}$.  Hence by part A (the Completeness Axiom),
there exists a least upper bound $L = \sup(S)$ in ${\bf R}$.  Let $\varepsilon > 0$
be any positive number.  By the definition of the least upper bound of a
set, $L - \varepsilon$ is not an upper bound for $S$.  Hence there exists $N$
such that $L - \varepsilon < x_N \le L$.  Since the sequence $(x_n)$ 
is monotone increasing, for all $n \ge N$, we have 
$L - \varepsilon < x_N \le x_n \le L$.  This implies that for 
all  $n \ge N$, $|x_n - L| < \varepsilon$.  Hence, 
$\lim_{n\to\infty} x_n = L$ by the definition of convergence
for a sequence.
\item{C)}  We are given that $x_n < 0$ for all $n\ge 1$.  Hence
from the inequality $x_{n+1}/x_n \le 1$ for all $n \ge 1$, we
get $x_{n+1} \ge x_n$ (the inequality ``flips'' when we multiply
by the negative number $x_n$).  Hence $(x_n)$ is monotone
increasing and bounded above by $0$.  By the theorem
from part B, $(x_n)$ converges to some $L \in {\bf R}$.  Note
that the most we can say about $L$ is: $L \le 0$.  $L$ could be
strictly negative too, though. 
\bigskip
\noindent
III.  
\item{A)}  We show $\lim_{n\to\infty} {4n \over 7n+5} = {4\over 7}$
using the definition.  Let $\varepsilon > 0$, and let 
$N = {20\over 49\varepsilon} - {5\over 7}$.  Then if $n > N$, we
have 
$$\eqalign{
\left|{4n \over 7n+5} - {4\over 7}\right| &= \left|{-20\over 7(7n+5)}\right|\cr
&= {20\over 7(7n+5)}\cr
&< {20\over 7(7\cdot\left({20\over 49\varepsilon} - {5\over 7}\right) + 5)}\cr
&= {20\over {20\over \varepsilon}}\cr
&=\varepsilon\cr}$$
Hence $\lim_{n\to\infty} {4n \over 7n+5} = {4\over 7}$.
\item{B)}  By the Limit Theorems for sequences,
$$\eqalign{
\lim_{n\to\infty} {4n^2 + 3n + 1\over 27n^3 + 3}
&=\lim_{n\to\infty} {4n^2 + 3n + 1\over 27n^3 + 3}\cdot {{1\over n^3}\over {1\over n^3}}\cr
&=\lim_{n\to\infty} {{4\over n} + {3\over n^2} + {1\over n^3}\over 27 + {3\over n^3}}\cr
&={\lim_{n\to\infty}{4\over n} + \lim_{n\to\infty}{3\over n^2} +\lim_{n\to\infty} {1\over n^3}\over 
27 + \lim_{n\to\infty}{3\over n^3}}\cr
&= {0\over 27}\cr
&=0\cr}$$
Note: The same sort of computation could be used to find the limit
for the sequence in part A. 
\bigskip
\noindent
IV.  
\item{A)}  For all $n$, we have by the triangle inequality,
$$|x_n| \le \left|\cos\left({\pi\sqrt{n}\over 2}\right) + {(-1)^n\over n}\right|
\le \left|\cos\left({\pi\sqrt{n}\over 2}\right)\right| + {1\over n} \le 1 + {1\over n} \le 2$$
Thus the sequence is bounded.
\bigskip
\item{B)}  If $\sqrt{n} = 4k$ for some integer $k$, or equivalently
$n = (4k)^2$, then
$\cos\left({4k \pi \over 2}\right) = \cos(2k\pi) = 1$ for all $k \ge 1$.  Hence
$$\lim_{k\to\infty} \left(\cos(2k\pi) + {(-1)^{(4k)^2}\over (4k)^2}\right) = 1 + 0 = 1.$$
So this subsequence converges to 1.  Similarly, if
$\sqrt{n} = 4k+1$ for some integer $k$, or equivalently
$n = (4k+1)^2$, then
$\cos\left({4k + 1 \pi \over 2}\right) = \cos\left(2k\pi + {pi\over 2}\right) = 0$ for all $k \ge 1$.  Hence
$$\lim_{k\to\infty} \left(\cos\left(2k\pi + {\pi\over 2}\right) + {(-1)^{(4k+1)^2}\over (4k+1)^2}\right) = 0 + 0 = 0.$$
So this subsequence converges to 0.  Since we have subsequences converging
to different limits, the sequence as a whole does not converge.
\bigskip
\noindent
{\it Extra Credit}
\bigskip
Since $(x_n)$ converges to $L$, we know that 
$(x_n)$ is bounded, hence there exists $M$ such that
$|x_n - L| \le M$ for all $n \ge 1$.  The idea of the 
proof is to split up $|y_n - L|$ into 
two groups of terms involving $|x_k - L|$ for $1 \le k \le n$.
In the first group, where $k$ is small, we use
the bound above.  For the others (with $k$ large),
we use the fact that $x_n$ is getting close to $L$
when $k$ is large.  The formal argument looks like this.
Let $\varepsilon > 0$.  Then there exists $N$ such that if $n > N$, then 
$|x_n - L| \le \varepsilon/2$.  Then for $n > N$, we have
$$\eqalign{|y_n - L| &= \left|{x_1 + \cdots + x_n \over n} - L\right|\cr
&\le {1\over n}(|x_1 - L| + \cdots + |x_N - L|) + {1\over n}(|x_{N+1} - L| + \cdots +|x_n - L|)\cr
&\le {N M\over n} + {(n-N)\varepsilon \over 2n}\cr}$$
The sequence $N M/n$ converges to zero as $n\to\infty$.  So given the same
$\varepsilon$ and $N$ fixed as above, there exists $N_1 \ge N$ such that 
$N M/n < \varepsilon/2$ for all $n > N_1$.  Then for all $n > N_1$, we have
$$|y_n - L| < \varepsilon/2 + {(n-N)\varepsilon \over 2n} < \varepsilon,$$
which shows what we want.
\bye
