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\begin{center}
{\bf Homework Set 5: Solutions}
\end{center}
\vspace*{.3in}

\noindent{\bf Page 78, \# 4.}

Note that for $n = 2m + 1$ odd, we have
$$
s_{2m+1} = \frac12 + s_{2m} = \frac{1+s_{2m-1}}{2}.
$$
I.e. $s_{2m+1}$ is the average of $s_{2m-1}$ and $1$.
So if we set $a_n = s_n - 1$, we get
$$
a_{2m+1} = \frac{1+a_{2m-1}+1}{2} - 1 = \frac{a_{2m-1}}{2}.
$$
Applying this formula repeatedly gives
$$
s_{2m+1} = 1+a_{2m+1} = 1 + \frac{a_1}{2^m} = 1 - \frac{1}{2^m}
$$
because $a_1 = s_1 - 1 = -1$.  Turning to the terms with even indices, we 
compute
$$
s_{2m} = \frac{s_{2m-1}}{2} = \frac{1}{2} - \frac{1}{2^m}.
$$

Now let $\{s_{n_k}\}_{k\in\N}$ be \emph{any} convergent subsequence with
limit $L$.  Then any further subsequence $\{s_{{n_k}_\ell}\}$ (!) must
converge to $L$, too.  So if $n_k$ is odd for infinitely many indices 
$k\in\N$, then the work above shows that $L = 1$.  And if this does not
happen, then $n_k$ must be odd for infinitely many indices, in which case
our work above shows that $L=1/2$.  All told, we see that any convergent
subsequence has limit equal to either 1 or $1/2$.  Therefore,
$$
\limsup s_n = 1,\qquad \liminf s_n = 1/2.
$$

\break
\noindent{\bf Page 80, \# 14}

\vspace*{.1in}
\noindent{\bf Part a: }
Given $\epsilon > 0$, choose $N_1\in\N$ such that $|s_n - s| < \epsilon/2$
for all $n\geq N_1$.  Set
$$
M = \max_{n<N_1} |s_n - s|
$$
and choose $N_2\in\N$ so that $N_1 M < (N_2+1)\epsilon/2$.  Then take
$N = \max\{N_1,N_2\}$.  If $n\geq N$, we estimate
\begin{eqnarray*}
|\sigma_n - s| & = & \frac{|s_0 + s_1 + \dots s_n - (n+1) s|}{n+1} \\ 
 	& \leq & \frac{|s_0 - s|}{n+1} + \dots \frac{|s_n - s|}{n+1} \\
	& < & \frac{|s_0 - s|}{n+1} + \dots \frac{|s_{N_1-1} - s|}{n+1} 
	+\frac{(n-N_1+1)}{n+1}\frac{\epsilon}{2} \\
	& < & N_1 \frac{M}{n+1} + \frac{n-N_1+1}{n+1}\frac{\epsilon}{2} \\
	& < & \frac{N_2+1}{n+1}\frac{\epsilon}{2}
	+ \frac{(n-N_1+1)}{n+1}\frac{\epsilon}{2} \\
	& < & \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.
\end{eqnarray*}
We conclude that $\lim_{n\to\infty}\sigma_n = s$.
\qed

\vspace*{.1in}
\noindent{\bf Parts b and c: }
Let $\{a_n\}$ be some sequence of positive numbers for which
$\sum a_n = m$ is finite.  Then in particular, $\lim a_n = 0$.
Let $s_n = k$ if $n=2^k$ for some $k\in\N$, but $s_n = a_n$ otherwise.
Then 
$$
\limsup s_n = \lim_{k\to\infty} s_{2^k} = \infty, \qquad
\liminf s_n = \lim a_n = 0
$$
so that $\lim_{n\to\infty} s_n$ does not exist.  Moreover, given
$n\in\N$, let $K$ be the largest integer such that $2^K\leq n$.  Then 
$$
\sigma_n < \frac{1}{n+1} \left(\sum_0^n a_n + \sum_{k=0}^K k\right)
	< \frac{1}{n+1}\left(m + \frac{K(K+1)}{2}\right)
	\leq \frac{m}{n+1} + \frac{K(K+1)}{2(2^K + 1)}
$$
which tends to zero as $n$ (and therefore $K$) tends to infinity. 

\break
\noindent{\bf Page 81, \# 16.}

\vspace*{.1in}
\noindent{\bf Part a: }
First note that if $x_n > \sqrt{\alpha}$, then 
$$
x_{n+1} - \sqrt{\alpha} = \frac{x_n^2 - 2x_n\sqrt{\alpha}+\alpha}{2x_n}
	= \frac{(x_n - \sqrt{\alpha})^2}{2x_n} > 0.
$$
so that $x_{n+1} > \sqrt{\alpha}$, too.  Since $x_1 > \sqrt{\alpha}$, 
we conclude that $x_n>\sqrt{\alpha}$ for every $n\in\N$.

In addition
$$
x_{n+1} - x_n = \frac{\sqrt{\alpha} - x_n}{2x_n} < 0,
$$
so that $\{x_n\}$ is a decreasing sequence that is bounded below by
$\sqrt{\alpha}$.  We conclude that $L := \lim_{n\to\infty} x_n$
exists and that $L\geq\sqrt{\alpha}$.

Taking the limit of both sides in the recursion formula for $x_n$ gives
$$
L = \frac{L^2 + \sqrt{a}}{2L}
$$
which, after rearranging, yields $L^2 = \alpha$.  Therefore $L=\sqrt{\alpha}$.
\qed

\vspace*{.1in}
\noindent{\bf Part b: }
The formula for $\epsilon_{n+1}$ in terms of $\epsilon_n$ is just
the first displayed equation in part (a).  Everything else follows
from $x_n > \sqrt{\alpha}$ and induction on $n$.

\vspace*{.2in}
\noindent{\bf Page 82, \# 20.}
Let $p = \lim_{i\to\infty} p_{n_i}$ be the limit of the convergent 
subsequence.  We will show using the definition of limit that
$p = \lim_{n\to\infty} p_n$.  To this end, let $\epsilon > 0$ be given.
By hypothesis, there exists $N_1\in\N$ such that $i\geq N_1$ implies that
$d(p_{n_i},p) < \epsilon/2$.  Also, since $\{p_n\}$ is Cauchy, there exists
$N_2\in\N$ such that $n,m\geq N_2$ implies $d(p_n,p_m) < \epsilon/2$.
Now let $N = \max\{N_1,N_2\}$ and choose $i \geq N$.  Then if $n\geq N$, we 
have
$$
d(p_n,p) \leq d(p_n,p_{n_i}) + d(p_{n_i},p) < \epsilon/2 + \epsilon/2 = 
	\epsilon.
$$
Note that the bound on the first term in the middle comes from the fact that
$n_i \geq i \geq N_2$, and the bound on the second term comes from the fact
that $i\geq N_1$.
\qed

\vspace*{.2in}
\noindent{\bf Page 82, \# 23.}
Let $d_n = d(p_n,q_n)$.  Then we must show that the sequence 
$\{d_n\}\subset\R$ converges.  Since $\R$ is complete, it will be enough
to show this sequence is Cauchy.  We prove this last fact directly from
the definition of Cauchy sequence.

Let $\epsilon > 0$ be given.  Then by hypothesis, there exists $N_1\in\N$
such that $d(p_n,p_m) < \epsilon/2$ for all $n,m\geq N_1$.  Similarly,
we have $N_2\in\N$ such that $d(q_n,q_m) <\epsilon/2$ for all 
$n,m\geq N_2$.  So taking $N = \max\{N_1,N_2\}$ and arbitrary
integers $n,m\geq N$, we apply the triangle inequality (four times!) 
to estimate
\begin{eqnarray*}
|d_n - d_m| & = & |d(p_n,q_n) - d(p_m,q_m)| \\
	    & = & |d(p_n,q_n) - d(p_n,q_m) + d(p_n,q_m) - d(p_m,q_m)| \\
	    & \leq & |d(p_n,q_n) - d(p_n,q_m)| + |d(p_n,q_m) - d(p_m,q_m)| \\
	    & \leq & d(q_n,q_m) + d(p_n,p_m) < \epsilon/2 + \epsilon/2 = 
	\epsilon.
\end{eqnarray*}
(To see what's going on with the second last inequality, it helps to 
draw yourself a picture.)
We conclude that $\{d_n\}$ is Cauchy, which is what we needed to prove.
\qed


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