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\noindent Solutions for Homework 5
\vskip .2in
  
\bigskip
\noindent {\bf Page 227/ \#2:}
\bigskip

\noindent{\it Part (a).} $\limsup a_n = 1$ and $\liminf a_n = 0$.
\medskip

\noindent{\it Part (b).} Let $b_n = a_{n+1}/a_n$.  Then 
$b_n = n!$ if $n = 2^k - 1$, $b_n = 1/n!$ if $n= 2^k$, and
$b_n = 1/(n+1)$ otherwise. Therefore, $\limsup b_n = \infty$.

\noindent{\it Part (c).} Let $b_n = a_n^{1/n}$.  Then $b_n = 1$ if
$n=2^k$ and $b_n = \sqrt[n]{1/n!}$ otherwise.  Since $b_n\leq 1$
for all $n$ and there are arbitrarily large $n$ (i.e. those $n$ equal
to $2^k$) for which $b_n = 1$, we conclude that $\limsup b_n = 1$.

\bigskip
\noindent {\bf Page 227/ \#6:}
\bigskip
\begin{eqnarray*}
\limsup (a_n + b_n) 
    & = & \lim_{n\to\infty} (\sup\{a_j + b_j\}_{j=n}^\infty)
          \leq \lim_{n\to\infty} (\sup\{a_j\}_{j=n}^\infty +
                            \sup\{b_j\}_{j=n}^\infty) \\
    & = & \lim_{n\to\infty}(\sup\{a_j\}_{j=n}^\infty)
          +\lim_{n\to\infty}(\sup\{b_j\}_{j=n}^\infty) 
          =\limsup a_n + \limsup b_n. 
\end{eqnarray*}
\qed

To see that the reverse inequality need not hold, try $a_n = 1 - (-1)^n$
and $b_n = (-1)^n$.  The sum of the $\limsup$'s is 3, but the $\limsup$
of the sum is 1.

\bigskip
\noindent {\bf Page 227/ \#8:}

Let $a = \limsup a_n$.  Let $j>0$ be an integer $\epsilon = 1/j$.  Then by 
Theorem 6.1, there exists $N\in\mathbf{N}$ such that $n\geq N$ implies that
$$
a_n - a < \epsilon.
$$
Also by Theorem 6.1, there exists {\it some} $n\geq N$, which we 
designate $n_j$ such that 
$$
a_{n_j} - a > -\epsilon.
$$
We find such an $n_j$ for each $j$ and have that
$$
-1/j = -\epsilon < a_{n_j} - a < \epsilon = 1/j
$$
for every $j$.  The squeeze theorem therefore guarantees that 
$\lim_{j\to\infty} a_{n_j} = a$.  Now technically, the $n_j$'s should
be increasing for $\{a_{n_j}\}$ to be a subsequence, but this can
be arranged by replacing $N$ with $N' = \max\{N,n_1,\dots,n_{j-1}\}$
when we go to choose $n_j$.  The result is then a subsequence $a_{n_j}$
converging to $a$ as $j$ goes to infinity.
\qed

\bigskip
\noindent{\bf Page 228/ \#9}
\bigskip

Again, let $a=\limsup a_n$.
Since each limit point of $\{a_n\}$ is the limit of a subsequence, 
the previous problem shows that $a\in P$.  Therefore, $\sup P\geq a$.  
On the other hand, if $c\in P$, then $c = \lim_{j\to\infty} a_{n_j}$
for some subsequence.  And if $\epsilon>0$, then there exists 
$N\in\mathbf{N}$ such that $a_n - a < \epsilon$ when $n\geq N$.
Since the $n_j$'s are increasing, there exists $J\in\mathbf{N}$ 
such that $n_j\geq N$ when $j\geq J$.  Therefore, $j\geq J$ implies
that 
$$
a_{n_j} - a < \epsilon.
$$
Letting $j$ tend to infinity shows that $c - a < \epsilon$.  
This is true for any $\epsilon>0$, so in fact, we see
that $c - a\leq 0$, i.e. $c\leq a$.  Hence $a$ is an upper bound
for $P$ and must therefore be at least as large as $\sup P$.
We conclude that $a=\sup P$.
\qed


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