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\centerline{Mathematics 242 - Principles of Analysis}
\centerline{Solutions - Exam 2}
\centerline{November 4, 2005}
\bigskip
\noindent
I. \item{A)} The Monotone Convergence Theorem says:  Every monotone bounded sequence 
in ${\bf R}$ is convergent.  Proof (in monotone increasing case):  
Say $(x_n)$ is a monotone increasing sequence that is bounded above.  
By the Axiom of Completeness, $a = \sup\{x_n : n\in {\bf N}\}$ exists in 
${\bf R}$.  By a result we know about least upper bounds, for all 
$\varepsilon > 0$, there exists some $N$ such that 
$a - \varepsilon < x_N \le a$.  But then since the sequence is monotone 
increasing, $a - \varepsilon < x_N \le x_n \le a$ for all $n \ge N$.  
It follows that $|x_n - a| < \varepsilon$ for all $n \ge N$, so by 
definition the sequence converges to $a$.
\item{B)}  Since $x_n < 0$, we have $x_{n+1} \ge x_n$ for all $n$.  
This means that the sequence is monotone increasing and bounded above by $0$.  
Hence part A shows that $\lim_{n\to\infty} x_n = a$ for some real number $a$.  
Furthermore $a\le 0$ by the Order Limit Theorem since $x_n < 0$ for all $n$.
\bigskip
\noindent
II.  \item{A)}  The Bolzano-Weierstrass theorem says that every bounded sequence has 
a convergent subsequence.
\item{B)}  We have $|x_n| = |\sin(n)| \le 1$ for all $n$.  Hence $x_n$ is a 
bounded sequence and the Bolzano-Weierstrass theorem applies.  There is a 
subsequence $x_{n_k}$ that converges.
\bigskip
\noindent
III. \item{A)} $\lim_{k\to\infty} ka_k = 4$ means that for $\epsilon = 1$, for 
instance, there exists a natural number $N$ such that $ka_k > 4 - 1 = 3$ 
for all $k > N$.  Then $a_k > {3\over k}$ for these
$k$, and the series $\sum_k a_k$ diverges by comparison with the series 
$\sum_k {3\over k}$.  So
{\it there are no such examples}.
\item{B)}  Any convergent series of negative terms is an example, for 
instance $\sum_{k=0}^\infty {-1\over 2^k}$.
\bigskip
\noindent
IV. \item{A)}  $\sum_{k=1}^\infty {1\over 5k+3}$ diverges by the comparison test:  
For all $k > 3$:
$5k + 3 < 6k$, so ${1\over 5k + 3} > {1\over  6k}$.  
The series $\sum_{k=1}^\infty  {1\over  6k}$
diverges because it is a constant times the harmonic series.   
However $\sum_{k=1}^\infty {(-1)^{k+1}\over 5k + 3}$ converges 
by the Alternating Series Test since ${1\over 5k + 3}$ is
strictly decreasing as 
$k \to \infty$ and $\lim_{k\to\infty} {1\over 5k + 3} = 0$.  The given series 
is {\it conditionally convergent}.
\item{B)}  This is a geometric series with ratio ${-1\over 7}$.  Since 
$|{-1\over 7}| < 1$ it is 
{\it absolutely convergent}.
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\noindent
V. Using the Ratio Test, 
$$\lim_{k\to\infty}\left|{(x-3)^{k+1}\over (k+1) 6^{k+1}}\cdot {k 6^k\over (x-3)^k}\right| 
= \lim_{k\to\infty} |x-3| \left({k+1\over 6 k}\right) = {|x-3|\over 6}$$
This is $<1$ when $|x - 3| < 6$, or $-3 < x < 9$.  The series converges 
absolutely for these $x$ and diverges for $x > 9$ and $x < -3$.  
When $x = 9$ we get the harmonic series which diverges.  When $x = -3$, 
then we get the alternating harmonic series which converges by the 
Alternating Series Test.  Hence the whole interval of convergence is 
$-3 \le x < 9$.  
\bigskip
\noindent
{\it Extra Credit.\/}  Let $\varepsilon > 0$.  Since $x_n$ converges to $a$ 
there exists a natural number $N_1$ such that $|x_n - a| < \varepsilon/2$ 
whenever $n \ge N_1$.  So when $n \ge N_1$, we
have by the triangle inequality 
$$\eqalign{
|y_n - a| &= \left|{x_1+x_2 +\cdots + x_n - na\over n}\right|\cr
& \le {1\over n}\left(|x_1 - a| + \cdots + |x_{N_1-1} - a| + |x_{N_1} - a| + \cdots + |x_n - a|\right)\cr
&\le   {1\over n}\left(|x_1 - a| + \cdots + |x_{N_1-1} - a|\right) + {n-N_1+1\over n}\varepsilon/2\cr
&< {1\over n}\left(|x_1 - a| + \cdots + |x_{N_1-1} - a|\right) + \varepsilon/2}$$
Now, using the Hint, notice that for all $k$, there exists a real number  $M$ such that 
$|x_k - a| \le M$ (a convergent sequence is bounded).  Therefore, the last 
term above is bounded above by ${N_1-1\over n}M$.  Given $\varepsilon$, 
$N_1$ is fixed as above.  But if we take $n$ sufficiently large, say 
$n > N_2$ for some $N_2$, then we can make 
${N_1-1\over n}M < \varepsilon/2$ also.  Therefore
$|y_n - a| < \varepsilon$ for all $n > \max(N_1,N_2)$.  
\bye

