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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 4}
\centerline{{\bf Due:} February 28, 2014}
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{\it `A' Section}
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1.  Using the `Big Theorem' on limits of sequences (Theorem 2.2.5 in the text), find
the limit of each of the following:
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\item{(a)}  $x_n = {(4^n + 9^{n/2})^2\over 2^{5n}}$
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\item{(b)}  $x_n = \sqrt{n}(\sqrt{25n + 3} - 5\sqrt{n})$ (Hint: 
$\sqrt{a} - \sqrt{b} = {a - b\over \sqrt{a} + \sqrt{b}}$ -- do you see why?  Also, you may use the result of question 1 in the B section if needed.)
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\item{(c)}  $x_n = {1\over n^4} \sum_{k=0}^4 {n\choose k} {1\over 3^k}$.  (The ${n\choose k}$ are the binomial coefficients as in Problem Set 2.)
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2.  For each of the following statements, first say whether the statement is true.  Then, if the 
statement is true, give a reason; if it is false give a counterexample.
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\item{(a)}  Let $\{x_n\}$ and $\{y_n\}$ be sequences.  If it is known that $\{x_n\}$ and 
$\{x_n - y_n\}$ converge, then $\{y_n\}$ converges too.
\item{(b)}  If $\{x_n\}$ and $\{y_n\}$ both diverge, then so does $\{x_n\cdot y_n\}$.
\item{(c)}  If $x_n \to 1/1000$, then there exists $n_0 \in {\bf N}$ such that $x_n > 0$
for all $n \ge n_0$.   
\item{(d)}  If $x_n > 0$ and ${x_{n+1}\over x_n} \le 1$ for all $n \ge 1$, then $x_n \to 0$.  
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3.  For each of the following sequences, first explain why the sequence is monotonic and bounded,
then determine the limit:
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\item{(a)}  $x_{n+1} = \sqrt{4 + x_n}$, $x_1 = 1$.
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\item{(b)}  $x_{n+1} = {3x_n + 7\over 8}$, $x_1 = 2$.  


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{\it `B' Section}
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\noindent
1.  Let $a > 0$.  Show, using the $\varepsilon,n_0$ definition,
that if $x_n \to a$ and $x_n \ge 0$ for all $n$, then $\sqrt{x_n} \to \sqrt{a}$.  
(Hint:  See part (b) of question 1 in the A section.)
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2. In this problem, you will show that the sequence $x_n = \sum_{k=1}^n {1\over k!}$ converges.
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\item{(a)}  First show that ${1\over k!} \le {1\over 2^{k-1}}$ for all integers $k \ge 1$.
\item{(b)}  Show that for all $n \ge 1$, 
$$1 + {1\over 2} + \cdots + {1\over 2^{n-1}} = {1 - {1\over 2^n} \over 1 - {1\over 2}}$$
and deduce that $x_n$ is bounded above by $2$.  
\item{(c)}  Use part (b) to show that $x_n$ converges.  
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3.  In class, we proved the Monotone Convergence Theorem for sequences, 
the result stated as Theorem 2.3.3 in the text.  In the proof we used the Least Upper Bound
Axiom in a crucial way.  It is an interesting fact that the implication goes the other
way too.  Namely if we assume that monotone increasing sequences of reals always converge
to real numbers, it follows that that every set of reals that is bounded above has
a least upper bound.  You will show this in this problem using the following construction.
Let $A$ be any nonempty set of real numbers that is bounded above.  The proof is based on 
constructing two sequences
$x_n$ and $y_n$ by this inductive procedure:
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\item{(i)}  Let $x_1 \in A$ be any element of $A$, and let $y_1$ be any upper bound for $A$.  
\item{(ii)}  Assuming $x_k$ and $y_k$ have been constructed such that $x_k \in A$ and
$y_k$ is an upper bound for $A$, 
find $x_{k+1}$ and $y_{k+1}$ like this:  Let $m_k = {x_k + y_k\over 2}$.  
If $m_k$ is also an upper bound for $A$, then set 
$x_{k+1} = x_k$ and $y_{k+1} = m_k$.  Otherwise, there is some $x \in A$ with $m_k < x \le y_k$, and 
in this case let $x_{k+1} = x$ and $y_{k+1} = y_k$.  
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\noindent
Note that an easy induction argument shows that this construction will produce
sequences such that $x_n \in A$ for all $n\ge 1$ and $y_n$ is an upper bound for $A$
for all $n\ge 1$.    
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Now prove the following statements to show that $A$ has a least upper bound in ${\bf R}$ assuming
the Monotone Convergence Theorem:
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\item{(a)}  Prove that both sequences $\{x_n\}$ and $\{y_n\}$ converge to elements of ${\bf R}$, using the Monotone Convergence Theorem and its Corollary.
\item{(b)}  Show that $\lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n$ by showing that 
$|x_n - y_n| = {|x_1 - y_1| \over 2^{n-1}}$ for all $n \ge 1$.  
\item{(c)}  Let $\alpha$ be the common limit of the $x_n$ and $y_n$ sequences.  Show that 
$\alpha = {\rm lub}(A)$ by showing it satisfies the definition of a least upper bound.
\bye    
