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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Final Exam Solutions -- December 14, 2005}
\bigskip
\noindent
I.  
\item{A)}  The sequence $(x_n)$ converges to $L$ if for every $\varepsilon > 0$,
there exists an $N \in {\bf N}$ such that $|x_n - L| < \varepsilon$
whenever $n \ge N$.
\item{B)}  
$$\lim_{n\to\infty} {5n^2 + 3n + 1\over n^2 + 4n} = \lim_{n\to\infty} {5 + {3\over n} +
{1\over n^2}\over 1 + {4\over n}} = 5$$
by the Algebraic Limit Theorem.
\item{C)}  Given $\varepsilon > 0$, let $N > {17\over \varepsilon}$.  Then
for all $n\ge N$, ${17\over n} < {17\over N} < \varepsilon$, so
$$\eqalign{
\left|{5n^2 + 3n + 1\over n^2 + 4n} - 5\right| &= {|-17n + 1|\over n^2+4n}\cr 
&= {17n - 1\over n^2 + 4n}\cr
&< {17n \over n^2}\cr
&= {17\over n} < \varepsilon\cr}$$
This shows the limit of the sequence is $5$ as claimed.
\bigskip
\noindent
II.  First we show $x_n$ is bounded above by 2 by induction.  The base
case is $n=1$, where $x_1 = 1 \le 2$ is clear.  Then assume $x_k \le 2$.
We have $x_{k+1} = \sqrt{x_k + 1} \le \sqrt{2 + 1} = \sqrt{3} \le 2$.  So
$x_n$ is bounded above by $2$ for all $n$.  

Next we show that $x_n$ is monotone increasing.  We have $x_2 = \sqrt{2} > 1 = x_1$
for the base case.  Next assume $x_k > x_{k-1}$.  Then 
$$x_{k+1} = \sqrt{x_k + 1} > \sqrt{x_{k-1} + 1} = x_k$$
So the sequence is monotone (strictly) increasing.

Finally, we know the sequence converges to some $L \le 2$ by the monotone
convergence theorem.  To find $L$ we take $\lim_{n\to\infty}$ in the
equation $x_n = \sqrt{x_n + 1}$, so $L = \sqrt{L + 1}$, or $L^2 - L - 1 = 0$.
This gives $L = {1 \pm \sqrt{5}\over 2}$ by the quadratic formula.  
The limit must be positive, so $L = {1 + \sqrt{5}\over 2}$.

\bigskip
\noindent
III.  The Bolzano-Weierstrass theorem is the statement that
every bounded sequence has a convergent subsequence.
See the class notes and text for the proof.

\bigskip
\noindent
IV.  
\item{A)}  The infinite series $\sum_{n=1}^\infty a_n$ converges if
and only if the sequence of partial sums: 
$S_N = \sum_{n=1}^N a_n$ converges to some $L\in {\bf R}$.
\item{B)}  This is a geometric series with first term $a = 15/7$ and
ratio $r = 3/7$, which is less than 1 in absolute value.  Hence the 
series converges to 
$${a\over 1-r} = {15/7 \over 1 - 3/7} = {15\over 4}.$$
\item{C)}  $f(n) = {1\over n + \ln(n)}$ is decreasing with $n$
and $\lim_{n\to\infty} {1\over n + \ln(n)} = 0$.  Hence the series
converges by the Alternating Series Test.  The series of absolute
values is $\sum_{n=1}^\infty {1 \over n + \ln(n)}$.  Since 
$\ln(n) < n$ for all $n \ge 1$, we have $n + \ln(n) < 2n$ and 
${1\over n + \ln(n)} > {1\over 2n}$.  The series $\sum_{n=1}^\infty {1\over 2n}$
diverges (it's a constant times the harmonic series).  Hence by the 
comparison test, $\sum_{n=1}^\infty {1\over n + \ln(n)}$
diverges too.  With the alternating signs, the series is {\it conditionally convergent}.
\item{D)}  
$$\eqalign{\lim_{n\to\infty} \left|{a_{n+1}\over a_n}\right| &=
\lim_{n\to\infty} \left|{(-1)^{n+1} (n+1)^2 \cdot 2^n\over (-1)^n n^2 2^{n+1}}\right|\cr
&= \lim_{n\to\infty} {1\over 2} \left(1 + {1\over n}\right)^2\cr
&= {1\over 2} < 1\cr}$$
Hence, the series converges absolutely.
\bigskip
\noindent
V.  
\item{A)}  No, $f(x)$ is not continuous at $x = 0$.  Let $x_n = {1\over n}$ (all 
rational.  Then $\lim_{n\to\infty} f(x_n) = \left({1\over n}\right)^2 - 1 = -1$.
But if $y_n = {\sqrt{2}\over n}$ (all irrational), then 
$\lim_{n\to\infty} f(y_n) = \lim_{n\to\infty} 2({\sqrt{2}\over n}) - 2 = -2.$
Since the two sequential limits are different, $f(x)$ is not continuous at 
$x = 0$, since $\lim_{x \to 1} f(x)$ does not exist. 
\item{B)}  Yes, $f(x)$ is continuous at $x = 1$.  Note that 
$$|f(x) - f(1)| = \cases{|x^2 - 1| = |x+1| |x-1| & if $x$ is rational\cr
                         2|x - 1| & if $x$ is irrational\cr}$$
Given $\varepsilon > 0$, let $\delta = \min\{1,\varepsilon/3\}$.  Then
if $|x - 1| < \delta$, we have $0 < 1 - \delta < x < 1 + \delta < 2$, so $1 < x+1  < 3$.
Hence in either case ($x$ rational or irrational), $|f(x) - f(1)| < 3\cdot \varepsilon/3 < \varepsilon$
or $|f(x) - f(1)| < 2\varepsilon/3 < \varepsilon$.  This shows that
$\lim_{x\to 1} f(x) = f(1)$ so $f$ is continuous at $x = 1$.
\bigskip
\noindent
VI.  
\item{A)}  This is TRUE.  By contraposition, if $f$ changes sign on the
interval, then the Intermediate Value Theorem shows $f(x) = 0$ for some $x \in [a,b]$. Hence
if $f(x)$ is never zero, then it is either always positive or
always negative.
\item{B)}  This is TRUE.  By the Extreme Value Theorem, $f$ attains
its minimum at some $c\in [a,b]$:  $f(x) \ge f(c)$ for all $x \in [a,b]$.
But we are given that $f(c) > 0$, so we can take $m = f(c)$.
\item{C)}  This is FALSE.  Counterexample:  $f : (-1,1) \to {\bf R}$
is continuous on the open interval, but the range $f(-1,1) = [0,1)$
which is not open.
\bigskip
\noindent
VII. By the definition,
$$\eqalign{
f'(0) &= \lim_{x\to 0} {f(x) - f(0) \over x - 0} \cr
&=\lim_{x\to 0} {x^3\sin(1/x) + 3x \over x}\cr
&=\lim_{x\to 0} x^2 \sin(1/x) + 3\cr
&=3\cr}$$
(Note $\sin(1/x)$ is bounded so $\lim_{x\to 0} x^2 \sin(1/x) = 0$.)
\bigskip
\noindent
VIII.  For the regular partition $P_n$ of $[1,3]$ we have $\Delta x = 2/n$
$x_i = 1 + 2i/n$ for $i = 0,\ldots,n$.  The function $f(x) = x^2 + 5$
is increasing on the interval $[1,3]$, so
$$U(f,P_n) = \sum_{i=1}^n \left(\left(1 + 2i/n\right)^2 + 5\right){2\over n}$$
and 
$$L(f,P_n) = \sum_{i=0}^{n-1} \left(\left(1 + 2i/n\right)^2 + 5\right){2\over n}$$
Hence $U(f,P_n) - L(f,P_n) = (14 - 6){2\over n} = {16\over n}$.  This can
be made $< \varepsilon$ by taking $n > 18\over \varepsilon$.  Hence
$f$ is integrable on $[1,3]$.  The value of the integral is
$$\eqalign{
\int_1^3 x^2 + 5\ dx &= \lim_{n\to\infty} U(f,P_n)\cr
&= \lim_{n\to\infty} \sum_{i=1}^n \left(\left(1 + 2i/n\right)^2 + 5\right){2\over n}\cr
&= \lim_{n\to\infty} \sum_{i=1}^n \left(6 + 4i/n + 4i^2/n^2 \right){2\over n}\cr
&= \lim_{n\to\infty} 12 + {8\over n^2}\sum_{i=1}^\infty i + {8\over n^3}\sum_{i=1}^\infty i^2\cr
&= \lim_{n\to\infty} 12 + {4n(n+1)\over n^2} + {4 n(n+1)(2n+1)\over 3n^3}\cr
&= 12 + 4 + 8/3 = 56/3}$$
\bigskip
\noindent
IX.
\item{A)}  This limit computes ${d\over dx} \ln(x)$ at $x = 1$.  By 
calculus we know ${d\over dx} \ln(x) = {1\over x}$, so 
$$\lim_{h\to 0} = {\ln(1+h)\over h} = 1.$$
\item{B)}  Writing $h = 1/n$, this shows
$$1=\lim_{n\to\infty} n\cdot \ln\left(1 + {1\over n}\right) = \lim_{n\to\infty} \ln\left(1 + {1\over n}\right)^n$$ 
Write $x_n = \ln\left(1 + {1\over n}\right)^n$.  Then the sequential criterion for continuity
shows 
$$\lim_{n\to\infty} \left(1 + {1\over n}\right)^n = \lim_{n\to\infty} e^{x_n}= e^{\lim_{n\to\infty} x_n} = e.$$

\bigskip
\noindent
{\it Extra Credit}.  Since $f$ is continous, it is integrable on 
$[1,B]$ for all $B > 1$.  Since $f$ is decreasing, the lower sum 
for the partition $P = \{1,2,\ldots, N\}$ on $[1,N]$ is the 
sum 
$$\sum_{n=2}^N f(n) = L(f,P) < \int_1^N f(x)\ dx < \lim_{B\to\infty}\int_1^B f(x)\ dx = L$$
Hence the partial sums of the series are bounded above by a fixed
number $L$.  By 
the Monotone Convergence Theorem, the partial 
sums converge to a finite limit, so $\sum_{n=1}^\infty f(n)$ converges.

To see that $\sum_{n=1} {1\over n^\alpha}$ converges for all
$\alpha > 1$, note that 
$$\lim_{B \to \infty} \int_1^B {1\over x^\alpha}\ dx = 
\lim_{B\to \infty} \left. {x^{-\alpha + 1}\over -\alpha + 1}\right|_1^B
= \lim_{B\to \infty} {B^{1-\alpha} - 1\over 1 - \alpha}$$
Since $\alpha > 1$, $1 - \alpha < 0$  so this limit exists, and equals
${1\over \alpha - 1}$.  Hence the sum converges by the first part
of the question. 


\bye
