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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 5}
\centerline{{\bf Due:} March 14, 2014}
\bigskip
\noindent
{\it `A' Section}
\bigskip
\noindent
1.  For each of the following sequences, determine three different subsequences,
each converging to a different limit.  For each one, express your three subsequences
as $x_{n_k}$ for a suitably chosen (strictly increasing) index sequence $n_k$, and give an explicit formula
for $n_k$ as a function of $k$:
\bigskip
\item{(a)}  $x_n = \sin\left({n \pi \over 2}\right)$
\medskip 
\item{} {\it Solution:}  If $n_k = 2k$, then $x_{n_k} = 0$ for all $k$, so that subsequence
converges to $0$.  If $n_k = 4k + 1$, then $x_{n_k} = 1$ for all $k$, so that subsequence 
coverges to $1$.  Finally, if $n_k = 4k + 3$, then $x_{n_k} = -1$ for all $k$ so that subsequence
converges to $-1$.  There are infinitely many other correct examples too.  
\medskip
\item{(b)}  $x_n = {n\over 5} - \left[{n\over 5}\right]$  (as usual, $[\ ]$ denotes the greatest integer function)
\medskip
\item{} {\it Solution:}  Let $n_k = 5k$, then $x_{5k} = k - k = 0$ for all $k$, so that subsequence 
converges to $0$.  Let $n_k = 5k + 1$, then $x_{5k+1} = k + 1/5 - k = 1/5$ for all $k$, so that 
subsequence converges to $1/5$.  Finally, if $n_k = 5k+2$, then the subsequence $x_{5k+2}$ converges
to $2/5$. (There are similar subsequences converging to $3/5$ and $4/5$ as well.)  
\bigskip
\noindent
2.  Let $x_n = \sqrt{n}$.  For each of the following sequences, either express that sequence
as a subsequence of
the sequence $x_{n_k}$ for some explicit (strictly increasing) index sequence $n_k$, or say why that is impossible:
\bigskip
\item{(a)}  $\{2,3,4,5,\ldots\}$
\medskip
\item{} {\it Solution:}  This is the subsequence $x_{n_k}$ for $n_k = (k+1)^2$, $k \ge 1$.  
\medskip
\item{(b)}  $\{\sqrt{3},\sqrt{6},\sqrt{9},\sqrt{12},\ldots\}$
\medskip
\item{} {\it Solution:}  This is the subsequence $x_{n_k}$ for $n_k = 3k$, $k \ge 1$.  
\medskip
\item{(c)}  $\{1,2,4,8,16,32,\ldots\}$
\medskip 
\item{} {\it Solution:}  This is the subsequence $x_{n_k}$ for $n_k = 2^{2(k-1)}$, $k \ge 1$.  
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\noindent
3.  Let $x_n = \sin\left({n\pi\over 4}\right)$ and $y_n = \cos\left({n\pi\over 4}\right)$.
\bigskip
\item{(a)}  Find a (strictly increasing) index sequence $n_k$ such that both $x_{n_k}$ and $y_{n_k}$ 
converge.
\medskip
\item{}  {\it Solution:}  The sequence $n_k = 8k + 1$ is such an index sequence since
$$\sin\left({(8k+1)\pi\over 4}\right) = \sin\left(2k\pi + {\pi\over 4}\right) = {\sqrt{2}\over 2}$$
and 
$$\cos\left({(8k+1)\pi\over 4}\right) = \cos\left(2k\pi + {\pi\over 4}\right) = {\sqrt{2}\over 2}$$
for all $k \ge 1$.  
\medskip
\item{(b)}  Find a second (strictly increasing) index sequence $n_k$ such that both $x_{n_k}$ and $y_{n_k}$ 
diverge.
\medskip
\item{} {\it Solution:}  The sequence $n_k = k$ is such a sequence.  
\medskip
\item{(c)}  Find a third (strictly increasing) index sequence $n_k$ such that one of $x_{n_k}$ and $y_{n_k}$ 
converges and the other diverges. 
\medskip
\item{} {\it Solution:}  The sequence $n_k = 4k$ is such an example since $\sin(k\pi) =0$ for 
all $k$, but $\cos(k\pi) = (-1)^k$ does not converge.  
\bigskip
\noindent
{\it `B' Section}
\bigskip
\noindent
1.  (True or False) -- If the statement is true give a proof; if it is false give a counterexample.
\bigskip
\item{(a)}  If $x_n$ is a sequence of strictly positive numbers converging to $0$, then 
$x_n$ has a strictly decreasing subsequence $x_{n_k}$.  
\medskip
\item{} {\it Solution:}  This is TRUE.  Here is one way to see it, constructing a strictly 
decreasing subsequence inductively.  We start with $n_1 = 1$, so $x_{n_1} = x_1$.  This is the 
base case.  Now assume we have found $x_{n_1} > x_{n_2} > \ldots > x_{n_k}$ for 
$1 = n_1 < n_2 < \cdots < n_k$.   Since $x_n \to 0$, 
given $\varepsilon = x_{n_k}$, there exists $N_0$ in ${\bf N}$ such that $x_n = |x_n - 0| < x_{n_k}$
for all $n \ge N_0$.  Take $n_{k+1} = \max(N_0,n_k + 1)$.  Then we have $x_{n_{k+1}} < x_{n_k}$, 
and $n_{k+1} > n_k$.  This shows we can continue to construct a strictly decreasing subsequence.     
\medskip
\item{(b)}  If $x_n \to 0$, then $x_n$ contains
a strictly increasing subsequence or a strictly decreasing subsequence (or both).  
\medskip
\item{} {\it Solution:}  This is FALSE.  Counterexamples are the constant sequence $x_n = 0$
for all $n\ge 1$, or any ``eventually constant'' sequence with $x_n = 0$ for all $n \ge n_0$
for some $n_0 \in {\bf N}$.  (Comment:  The statement would be true if we assumed $x_n \ne 0$ for
all $n$ (or even all $n\ge n_0$ for some $n_0 \in {\bf N}$).  
Any such sequence contains either infinitely many positive terms or
infinitely many negative terms, or both.  If there are infinitely many positive terms, we get
a strictly decreasing subsequence of the positive terms by part (a) of this question. 
If there are infinitely many strictly negative terms, then there is a strictly increasing
subsequence, as can be seen by taking negatives, using part (a), then flipping signs
again.)
\medskip 
\item{(c)}  If $x_n$ is a monotone increasing sequence with a bounded subsequence $x_{n_k}$, then 
$x_n$ converges.
\medskip
\item{} {\it Solution:}  This is TRUE.  Since the whole sequence is increasing, so is the 
subsequence $x_{n_k}$.  But that subsequence is bounded above, say by $a \in {\bf R}$.  
We claim that the whole sequence $x_n$ is also bounded above by $a$.  To see
that, let $n$ be any natural number.  Since $\{n_k\}$ is a strictly increasing sequence of natural numbers, it follows that it is not bounded above.  So there is some $k$ such that $n_k \ge n$.  But 
then since $x_n$ is increasing and $a$ is an upper bound for the subsequence, 
$a \ge x_{n_k} \ge x_n$.  It follows that $a \ge x_n$ for all $n$, and hence
the whole sequence is bounded above.  
Then $\{x_n\}$ converges as well by the Monotone Convergence Theorem.   
(You can also show that the limit of the whole sequence must be the same as the limit
of the subsequence, but that was not required.)

\bigskip
\noindent
2.  Consider the sequence $x_n = \cos(n)$ (where we think of $n$ as an angle expressed in radians).
\bigskip
\item{(a)}  Prove that $x_n$ has a convergent subsequence.
\medskip
\item{} {\it Solution:}  Since $|\cos(n)| \le 1$ for all $n\ge 1$, this is a bounded sequence.  The
statement to be proved is a direct consequence of the Bolzano-Weierstrass Theorem. 
\medskip
\item{(b)}  In this part of the question we will show that $x_n$ is not convergent, though.
Suppose $\lim_{n\to\infty} \cos(n) = a$ for some real number $a$.  Using a trig identity for 
$\cos(n + 1)$ and considering $\lim_{n\to\infty} (\cos(n+1) - \cos(n))$, show that 
$${a (\cos(1) - 1)\over \sin(1)} = \lim_{n\to\infty} \sin(n).$$  But then use the sequence
$\lim_{n\to\infty} (\sin(n+1) - \sin(n))$ to deduce that $a = 0$, so 
$\lim_{n\to\infty} \cos(n) = \lim_{n\to\infty} \sin(n) = 0$.  But this is a contradiction.
Explain why to conclude the proof.  
\medskip
\item{} {\it Solution:}  The addition formula for $\cos$ implies that 
$\cos(n + 1) = \cos(n)\cos(1) - \sin(n)\sin(1)$.  If we assume that $\lim_{n\to\infty} \cos(n) = a$, then
using parts of the ``Big Theorem'' and rearranging algebraically, we see
$$\eqalign{
0 &= \lim_{n\to\infty} (\cos(n+1) - \cos(n)) \cr
&= \lim_{n\to\infty} \cos(n)(\cos(1) - 1) - \lim_{n\to\infty} \sin(n) \sin(1)\cr
&= a(\cos(1) - 1) - \sin(1) \lim_{n\to\infty} \sin(n).\cr}$$
Thus, 
$$\lim_{n\to\infty} \sin(n) = {a(\cos(1) - 1)\over \sin(1)}. \leqno(1) $$
as claimed.  Now the addition formula for $\sin$ shows 
$$\sin(n+1) = \sin(n) \cos(1) + \cos(n) \sin(1)$$
Taking the limit as $n \to \infty$ on both sides and substituting from (1), we get:
$${a(\cos(1) - 1)\over \sin(1)} = {a(\cos(1) - 1)\over \sin(1)} \cos(1) + a\sin(1),$$
so 
$$a(\cos(1) - 1) = a(\cos(1) - 1)\cos(1) + a\sin^2(1),$$
and hence (because $\cos^2(1) + \sin^2(1) = 1$), 
$$a(\cos(1) - 1) = a(1 - \cos(1)).$$
The only way this can be true is if $a = 0$.  But then this implies that 
$$\lim_{n\to\infty} \sin(n) = 0 = \lim_{n\to\infty} \cos(n).$$
But that is impossible because it would say $\lim_{n\to\infty} \cos^2(n) + \sin^2(n) = 0$ by
parts (a) and (b) of the ``Big Theorem.''  However, we know by the basic identity for the 
trigonometric functions that  
$$\cos^2(n) + \sin^2(n) = 1$$ 
for all $n\ge 1$.  Hence
$\lim_{n\to\infty} \cos^2(n) + \sin^2(n) = 1$ as well if the two limits exist.   This 
contradiction shows that $\lim_{n\to\infty} \cos(n)$ cannot exist.  
\bigskip
\noindent
3.  A {\it cluster point} of a sequence $x_n$ is a limit of a convergent subsequence $x_{n_k}$.
(See question 1 on the A section for examples of sequences with several different cluster points.)
Show that if $a_m$ is a convergent sequence of cluster points of a given sequence $x_n$,
then $a = \lim_{m\to\infty} a_m$ is also a cluster point of the $x_n$ sequence.  
\medskip
\item{} {\it Solution:}  We let $a_m$ be a convergent sequence of cluster points of $x_n$ with 
$a = \lim_{m\to\infty} a_m$.  Since each $a_m$ is a cluster point of the $x_n$ sequence, there
is a subsequence of $x_n$ converging to $a_m$.  From the subsequence converging to $a_1$,
select any $x_{n_1}$ with $|x_{n_1} - a_1| < 1$. Then from the subsequence converging 
to $a_2$, select any $x_{n_2}$ with $n_2 > n_1$ and $|x_{n_2} - a_2| < {1\over 2}$, then 
from the subsequence converging to $a_3$, select $x_{n_3}$ with $|x_{n_3} - a_3| < {1\over 3}$
and $n_3 > n_2$.  By an induction argument, we can always continue this process since for 
any $\ell\ge 1$, there are infinitely many index values $n_\ell$ for which 
$|x_{n_\ell} - a_\ell| < {1\over \ell}$
(all the indices giving terms in the subsequence converging to $a_\ell$ that are 
distance ${1\over \ell}$ or less from $a_\ell$).    
In this way we get a subsequence $\{x_{n_\ell}\}$ (indexed by $\ell \in {\bf N}$)
with $n_\ell$ strictly increasing and $|x_{n_\ell} - a_\ell| < {1\over \ell}$
for all $\ell \ge 1$.  We claim that the subsequence $x_{n_\ell}$ converges to $a$.  
To see this note that given any $\varepsilon > 0$, there exists $\ell_0 \in {\bf N}$ 
such that ${1\over \ell} < {\varepsilon \over 2}$ and $|a_\ell - a| < {\varepsilon\over 2}$ 
for all $\ell \ge \ell_0$.  Then by the triangle inequality, for all $\ell \ge \ell_0$,
$$|x_{n_\ell} - a| = |x_{n_\ell} - a_\ell + a_\ell - a| \le |x_{n_\ell} - a_\ell| + |a_\ell - a| < {\varepsilon\over 2} + {\varepsilon\over 2} = \varepsilon.$$
This shows that the subsequence $\{x_{n_\ell}\}$ converges to $a$, so $a$ is also a cluster point of the $\{x_n\}$ sequence.   

\bye    
