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\n {\bf \textcolor{red}{Note:}} \textcolor{blue}{The
homework you
turn in must contain each problem statement in its entirety
and
followed by its solution, as demonstrated in the first
problem
below. Others below are sketches, outlines, or hints for
solutions.}
\n {\bf \textcolor{red}{[\#1.3]}} Prove
$1^3+2^3+\cdots+n^3=(1+2+\cdots+n)^2$ for all natural
numbers $n$.
\noindent Proof: Notice first that
$1+2+\cdots+n=\frac{n(n+1)}{2}$. So we only need to show
$1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$. By induction, we
have
$1^3=\frac{1^2(1+1)^2}{4}$ for $n=1$. So the identity holds
for
$n=1$. Assume it holds for $n$. Now consider the case for
$n+1$.
By the induction assumption, we have
$1^3+2^3+\cdots+n^3+(n+1)^3=\frac{n^2(n+1)^2}{4}+(n+1)^3=(n+1)^2[\frac{n^2}{4}+(n+1)]
=(n+1)^2\frac{n^2+4n+4}{4}=\frac{(n+1)^2(n+2)^2}{4}$, which
is the
case for $n+1$. This completes the proof. $\square$
\n {\bf \textcolor{red}{[\#1.4}} (a) Let
$S_n=1+3+5+\cdots+(2n-1)$. Then $S_1=1,S_2=4,S_3=9$,
suggesting
$S_n=n^2$. (b) Assume $S_n=n^2$. Then
$S_{(n+1)}=S_n+2(n+1)-1=n^2+2n+1$ by the assumption and
simplification, which equals $(n+1)^2$ by complete squaring.
Hence, by induction we have shown that $S_n=n^2$ for all
$n\in{\Bbb N}$.
\n {\bf \textcolor{red}{[\#1.6 *]}} Modelled after Example 2
page
3.
\n {\bf \textcolor{red}{[\#1.12*]}} (a) Notice first
that for all
$n\in{\Bbb N}$,
$\left(\begin{array}{c}n\\
0\end{array}\right)=\frac{n!}{0!(n-0)!}=1,
\left(\begin{array}{c}n\\
n\end{array}\right)=\frac{n!}{n!(n-n)!}=1$,
$n!=n(n-1)\cdots 2\cdot1=n(n-1)!=n(n-1)(n-2)!$,
$\left(\begin{array}{c}n\\
1\end{array}\right)=\frac{n!}{1!(n-1)!}=n,
\left(\begin{array}{c}n\\
2\end{array}\right)=\frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}$.
With
those identities, you can check the cases for $n=1,2,3$
directly.
Also, notice that there are $n+1$ terms in the binomial
expansion
$(a+b)^n$. (b) $\left(\begin{array}{c}n\\
k\end{array}\right)+\left(\begin{array}{c}n\\
k-1\end{array}\right)=\frac{n!}{k!(n-k)!}+\frac{n!}{(k-1)!(n-(k-1))!}$
Because $k!=k(k-1)!, (n-(k-1))!=(n+1-k)!=(n+1-k)(n-k)!$, the
common denominator is $k(k-1)!(n+1-k)(n-k)!=k!(n+1-k)!$.
Simplify
the addition then as follows: $\left(\begin{array}{c}n\\
k\end{array}\right)+\left(\begin{array}{c}n\\
k-1\end{array}\right)=\frac{n!(n+1-k)+n!k}{k!(n+1-k)!}=\frac{n![(n+1-k)+k]}{k!(n+1-k)!}
=\frac{n!(n+1)}{k!(n+1-k)!}=\frac{(n+1)!}{k!(n+1-k)!}=\left(\begin{array}{c}n+1\\
k\end{array}\right)$. (c) Use induction. More precisely, the
expansion for $n=1$ trivially. Assume the expansion for $n$.
Then
consider the case for $n+1$:
$(a+b)^{n+1}=(a+b)^n(a+b)=\left[\left(\begin{array}{c}n\\
0\end{array}\right)a^n+\left(\begin{array}{c}n\\
1\end{array}\right)a^{n-1}b+\cdots\left(\begin{array}{c}n\\
n\end{array}\right)b^n\right](a+b)$, by the assumption. Expand
further, collect like terms $a^{n+1}, a^nb,\dots,
a^kb^{n-k},
b^{n+1}$. You will find with exceptions for the 1st and last
terms, the coefficient for $a^kb^{(n-k)}$ is
$\left(\begin{array}{c}n\\
k\end{array}\right)+\left(\begin{array}{c}n\\
k-1\end{array}\right)=\left(\begin{array}{c}n+1\\
k\end{array}\right)$ by (b) for $k=1,2,\dots, n$. For
$a^{n+1},
b^{n+1}$, their
coefficients remain to be $\left(\begin{array}{c}n\\
0\end{array}\right)=\frac{n!}{0!(n-0)!}=\left(\begin{array}{c}n+1\\
0\end{array}\right)=1, \left(\begin{array}{c}n\\
n\end{array}\right)=\frac{n!}{0!(n-0)!}=\left(\begin{array}{c}n+1\\
n+1\end{array}\right)=1$.
\n {\bf \textcolor{red}{[\#2.2, 2.4, 2.5]}} They are all
similar.
Follow Examples 2--6 of \S 2.
\n {\bf \textcolor{red}{[\#3.3]}}
\begin{equation*}
\begin{split}
(-a)(-b) &=(-a)(-b)+0\
(\hbox{by A3})\\
&=(-a)(-b)+(ab+(-ab))\ (\hbox{A4})\\
&=[(-a)(-b)+(-ab)]+ab\ (\hbox{A2, A1})\\
&=[(-a)(-b)+(-a)b)]+ab\ (\hbox{Thm 3.1(iii)})\\
&=(-a)[(-b)+b)]+ab\ (\hbox{DL})\\
&=(-a)0+ab=ab\ (\hbox{A4, A3})
\end{split}
\end{equation*}
\n {\bf \textcolor{red}{[\#3.5]}} (a) First it is obvious
that
$-|b|\le b\le |b|$ because either $|b|=b$ or $|b|=-b$ which
implies for the former case that $|b|=b\ge 0\ge -|b|$ and
that
$-|b|=b\le 0\le |b|$ for the latter case. Therefore,
together with
$|b|\le a$ it implies $-a\le -|b|\le b\le |b|\le a$ as
required.
Conversely, if $-a\le b\le a$, then we must have $|b|=b\le
a$ if
$b\ge 0$ using the right part of the inequality $b\le a$, or
$|b|=-b\le a$ if $b\le 0$ using the left part of the
inequality
$-a\le b\Rightarrow -b\le a$. (b) By (a) we only need to
show
$-|a-b|\le |a|-|b|\le |a-b|$. For the right part, we have
$|a|=|a-b+b|\le |a-b|+|b|$ by the triangle inequality.
Hence,
$|a|-|b|\le |a-b|$. This is true for all $a,b$. Exchanging
$a, b$,
we have $|b|-|a|\le |b-a|=|-(b-a)|=|a-b|$ which is the same
as
$-|a-b|\le |a|-|b|$, showing the left part of the
inequality.
\n {\bf \textcolor{red}{[\#3.6]}} (a) $|a+b+c|=|a+(b+c)|\le
|a|+|b+c|\le |a|+|b|+|c|$, using the triangle inequality
twice in
a roll. (b) Follow the hint.
\n {\bf \textcolor{red}{[\#3.7*]}} (a) Same as 3.5(a)
changing
$\le$ to $<$ in the argument. Alternatively, it is a
special case
of 3.5(a). More precisely, for $|b|<a$, we cannot have
$b=a$ if
$b\ge 0$ nor $b=-a$ since $-a<-|b|=b$. Thus, $|b|<a$
iff $-a\le
b\le a$ with $b\ne a, -a$ iff $-a<b<a$. (b) By (a),
$|a-b|<c\Longleftrightarrow -c<a-b<c
\Longleftrightarrow
b-c<a<b+c$. (c) Same as (b), changing $<$ to $\le$
in the
argument. Alternative, consider the two cases $|a-b|<c$
and
$|a-b|=c$ separately. For the former,
$|a-b|<c\Longrightarrow
b-c<a<b+c$ by (b), $\Longrightarrow b-c\le a\le b+c$.
Conversely,
$|a-b|<c$ implies $a-b\ne c, -c$, which together with
$b-c\le a\le
b+c$ implies
$b-c< a< b+c \Longrightarrow |a-b|<c$ by (b). For
the latter case that $|a-b|=c$, either $a-b=c$ or $-c$,
implying
$a=b+c$ or $b-c$, implying $b-c\le a\le b+c$. Conversely,
$b-c\le
a\le b+c\Longrightarrow -c\le a-b\le c$. Together with the
case
definition $|a-b|=c$ we have the trivial conclusion
$|a-b|\le c$.
\n {\bf \textcolor{red}{[\#3.8*]}} Assume instead that
$a>b$. Then
$b<2^{-1}(a+b)<a$ because
$2b=b(1+1)=b+b<a+b<a+a=2a$. Let
$b_1=2^{-1}(a+b)>b$. Then by the hypothesis we have $a\le
b_1=2^{-1}(a+b)\rightarrow
2a=2\cdot2^{-1}(a+b)=a+b\rightarrow
a\le b$, contradicting the assumption that $a>b$.
\n {\bf \textcolor{red}{[\#4.5]}} Since $s\le m=\sup S$ for
all
$s\in S$ and $m\in S$, therefore by definition $m=\max S$.
\n {\bf \textcolor{red}{[\#4.6]}} (a) Since $S\ne
\emptyset$,
$\exists s_0\in S$
s.t. $\inf S\le s_0\le \sup S$. (b) $S$ must
be a one-point set $S=\{a\}$ for some $a\in{\Bbb R}$.
\n {\bf \textcolor{red}{[\#4.7*]}} (a) $\forall s\in
S\subset T$,
$\inf T\le s$ by a part of the definition of $\inf T$. This
implies $\inf T$ is a lower bound of $S$. By $\inf S$, we
must
have $\inf T\le \inf S$. You then show similarly that $\sup
S\le
\sup T$. The part $\inf S\le \sup S$ is from \#4.6. (b)
Since $S,
T\subset S\cup T$, by (a), $\sup S, \sup T\le \sup(S\cup T)$
and
$\max\{\sup S,\sup T\}\le \sup(S\cup T)$. One the other
hand,
$\forall a\in S\sup T$, either $a\in S$ or $a\in T$, which
implies
either $a\le \sup S$ or $a\le \sup T \Longleftrightarrow
a\le\max\{\sup S,\sup T\}$. Therefore, $\max\{\sup S,\sup
T\}$ is
an upper bound of $S\cup T$. Because $\sup(S\cup T)$ is the
least
upper bound, $\sup(S\cup T)\le \max\{\sup S,\sup T\}$.
Together
the established inequality
$\max\{\sup S,\sup T\}\le \sup(S\cup
T)$ we have the equality $\max\{\sup S,\sup T\}=\sup(S\cup
T)$.
\n {\bf \textcolor{red}{[\#4.10*]}} By Archimedean Property,
$\exists k\in {\Bbb N}$ s.t. $ka>1$ since $a>0,
1>0$. Thus
$a>\frac{1}{k}$ since $k>0$. Use the property for same
pair $1,
a$, $\exists m\in{\Bbb N} \Rightarrow m=m\cdot 1>a$. Let
$n=\max{k,m}$, then $\frac{1}{n}\le \frac{1}{k}<a<m\le
n$.
\n {\bf \textcolor{red}{[\#4.11]}} It suffices to show there
is an
infinite sequence
$a<a_1<a_2<\dots<a_n\dots<b$ with $a_n\in{\Bbb
Q}$. Construct the sequence by induction. By the denseness
of
$\Bbb Q$, $\exists a_1\in {\Bbb Q} \Rightarrow
a<a_1<b$. Assuming
$a_n$ is constructed such that $a<a_1<a_2<\dots a_n
<$, then
applying the same denseness property to the pair $a_n<b$
to have
some $a_{n+1}\in {\Bbb Q}$ with $a_n<a_{n+1}<b$. This
completes
the proof.
\n {\bf \textcolor{red}{[\#4.12]}} By the denseness property
of
$\Bbb Q$ in $\Bbb R$, we have for this pair $a-\sqrt
2<b-\sqrt 2$
a rational $r\in {\Bbb Q}$ such that $a-\sqrt
2<r<b-\sqrt 2$,
which is $a<r+\sqrt 2<b$. Since $r\in {\Bbb Q}, \sqrt
2\in {\Bbb
I}$, we must $x=r+\sqrt 2\Bbb I$ for otherwise $\sqrt
2=x-r\in\Bbb
Q$ would be a contradiction.
\n {\bf \textcolor{red}{[\#4.16*]}} By the way set
$A:=\{r\in{\Bbb
Q}:r<a\}$ is defined, we concluded right away that $a$ is
an upper
bound of $A$: $\sup A\le a$. If $a\ne \sup A$, then we must
have
$\sup A<a$. By the denseness property of $\Bbb Q$, there
is a
$r\in \Bbb Q$ such that $\sup A<r<a$. By definition of
$A$, this
$r\in A$ contradicting the implication that $a\le \sup A$.
\n {\bf \textcolor{red}{[\#5.2]}}
\n {\bf \textcolor{red}{[\#5.4*]}} Consider 2 cases
separately.
Case of $m=\inf S>-\infty$. Then $\forall s\in S$, $m\le
s$ which
implies $-s\le -m$. Thus $-m$ is an upper bound for $-S$.
Moreover, if $A$ is an upper bound of $-S$: $-s\le A \forall
s\in
S$, then $-A\le s \forall s\in S$ is an lower bound of $S$,
and
$-A\le m=\inf S$ follows. Thus $-m\le A$, and $-m=\sup(-S)$
by
definition and $\inf S=m=-(-m)=-(\sup(-S))$ follows. For the
remaining case that $\inf S=-\infty$ that $S$ is not bounded
below, then $-S$ cannot be bounded above because $M$ is an
upper
bound of $-S$ iff $-M$ is a lower bound of $S$. Hence
$\sup(-S)=\infty$, and $\inf S=-\infty=-\sup(-S)$.
\n {\bf \textcolor{red}{[\#5.5]}} The argument is identical
to
4.6(a).
\n {\bf \textcolor{red}{[\#7.2]}}
\n {\bf \textcolor{red}{[\#7.4*]}} (a) $\sqrt 2/n\to 0$. (b)
$(1+1/n)^n\to e$. $t_{n+1}=(t_n^2+2)/(2t_n), t_1=1$. Then
$1\le
t_n\le 2$, and $t_n$ increasing. $\lim t_n=t$ exists, and
$t=\sqrt
2$.
\n {\bf \textcolor{red}{[\#8.2a,c]}} (a) $\forall
\epsilon>0$ let
$N=1/\epsilon$. Then $n>N\Longrightarrow
|a_n-0|=a_n=n/(n^2+1)<n/n^2=1/n<1/N=\epsilon$.
\n {\bf \textcolor{red}{[\#8.4]}} By assumption,
$\forall\epsilon>0, \exists N>0\ s.t.\ n>N\ \Longrightarrow \
|s_n|<\epsilon/M,\Longrightarrow
|s_nt_n|=|s_n||t_n|<(\epsilon/M)M=\epsilon$ since
$|t_n|<M$ for
all $n$.
\n {\bf \textcolor{red}{[\#8.8(a)*]}} \textcolor{blue}{(This
is
another example for how YOUR hand-in homework should look
like for
this problem: State the problem, followed by a formal
declaration
``Proof'' or ``Solution'' whichever applies.)}
\n
Prove the limit $\lim(\sqrt{n^2+1}-n)=0$.
\n Proof: $\forall \epsilon>0$, let $N=1/\epsilon$. Then
for $n>N$
we have
$|\sqrt{n^2+1}-n-0|=\sqrt{n^2+1}-n=(\sqrt{n^2+1}-n)\cdot(\sqrt{n^2+1}+n)/(\sqrt{n^2+1}+n)=
(\sqrt{n^2+1}^2-n^2)/(\sqrt{n^2+1}+n)=1/(\sqrt{n^2+1}+n)<1/n<1/N=\epsilon$.
This proves $\lim(\sqrt{n^2+1}-n)=0$ by definition.
$\square$
\n {\bf \textcolor{red}{[\#9.2b]}} By Theorems 9.2 and 9.3,
$\lim(3y_n-x_n)=\lim(3y_n+(-1)x_n)=3\cdot 7+(-1)\cdot 3=18$.
By
Theorem 9.6, $\lim(3y_n-x_n/y_n=18/7$.
\n {\bf \textcolor{red}{[\#9.4]}} (b) $s_1=1,
s_{n+1}=\sqrt{s_n+1}$. Assume $\lim s_n=s$ exists. Than
$\lim
s_{n+1}=\lim s_n=s$. By Example 5 of $\S$ and Theorem 9.3,
$s=\lim
s_{n+1}=\lim\sqrt{s_n+1}=\sqrt{\lim s_n+1}=\sqrt{s+1}$.
Solving
$s$ we get $s=(1+\sqrt 5)/2$.
\n {\bf \textcolor{red}{[\#9.6a,b]}} (a) Plug in the
``limit'' to
get $a=3a^2\Longrightarrow a=0 \text{ or } a=1/3$. (b) The
limit
does not exits because $x_n>3^{n-1}\to\infty $ by
induction.
\n {\bf \textcolor{red}{[\#9.8]}}
\n {\bf \textcolor{red}{[\#9.10*]}} (a) By assumption,
$\forall
M>0, \exists N,\ s.t.\ n>N\Longrightarrow
s_n>M/k>0$ since $k>0$
is a constant. Therefore $ks_n>k(M/k)=M$ for all
$n>N$, showing
$ks_n\to\infty$ by definition. (b) $(\Longrightarrow)$
$\forall
M<0,\ \exists N,\ s.t. \ n>N\Longrightarrow s_n>-M$
since $\lim
s_n=+\infty$. Hence we have $-s_n<M$ showing
$\lim(-s)=-\infty$ by
definition. Similar argument applies to $(\Longleftarrow)$.
Also
for (c).
\n {\bf \textcolor{red}{[\#9.12*]}} (a) Let
$\epsilon_0=(1-L)/2>0$
as $L<1$ . By assumption $\exists N_0>0$ such that
$\forall n\ge
N_0, ||s_{n+1}|/|s_n|-L|<\epsilon_0 \Longleftrightarrow
L-\epsilon_0<|s_{n+1}|/|s_n|<L+\epsilon_0=(1+L)/2$.
Let
$a=(1+L)/2$. Then $L<a<1$, and $|s_{n+1}|/|s_n|<a$
for $n\ge N_0$.
Repeatedly using this inequality for $n, n-1, \dots,
n-(n-N_0)=N_0>\ge N_0$, we have
$|s_n|<a|s_{n-1}|<a(a|s_{n-2}|)<\dots
<a^{n-N_0}|s_{n-(n-N_0)}|=a^{n-N_0}|s_{N_0}|$. Because
$N_0$ is
fixed and $a^n\to 0$ as $n\to \infty$ since $0<a<1$,
we have
$\forall \epsilon>0, \exists N$ s.t.
$n>N\Longrightarrow
a^n<\epsilon a^{N_0}/|s_{N_0}|\Longrightarrow
|s_n|<a^{n-N_0}|s_{N_0}|<\epsilon$. (b) Let
$t_n=1/|s_n|$. Then
$t_{n+1}/t_n=1/(s_{n+1}/s_{n})\to 1/L<1$. By (a) $\lim
t_n=0$
which is equivalent to $s_n\to\infty$ by Theorem 9.10.
\n {\bf \textcolor{red}{[\#9.14]}} Follow the hints.
\n {\bf \textcolor{red}{[\#9.16]}} Follow the instruction.
\n {\bf \textcolor{red}{[\#10.6*]}} (a) $\forall
\epsilon>0$, let
$N=-\frac{\ln\epsilon}{\ln 2}$, then $n\ge m> N$ implies
\begin{equation*}
\begin{split}
|s_n-s_m|&=|s_n-s_{n-1}+s_{n-1}-s_{n-2}+\dots-s_{m+1}+s_{m+1}-s_m|\\
&\le
|s_n-s_{n-1}|+|s_{n-1}-s_{n-2}|+\cdots+|s_{m+1}-s_m|\\
&<2^{-(n-1)}+2^{-(n-2)}+\cdots+2^{-(m+1)}+2^{-m}\\
&=2^{-m}\frac{1-2^{-(n-m)}}{1-2^{-1}}\\
&<2^{-(m-1)}\le 2^{-N}=\epsilon.\\
\end{split}
\end{equation*}
(b) No. Conterexample: $s_n=\sum_{k=1}^n1/k\to\infty$ as
$n\to\infty$ hence, it cannot be Cauchy for every Cauchy
sequence
must be bounded. However $s_{n+1}-s_n=1/(n+1)<1/n$
satisfied.
\n {\bf \textcolor{red}{[\#10.10*]}} (a, b) are
straightforward.
(c) Assume $(s_n)$ is not nonincreasing, then there is an
$n$ such
that $s_{n+1}>s_n\Longleftrightarrow (s_n+1)/3>s_n
\Longleftrightarrow s_n<1/2$ contradicting $s_n\ge 1/2$
for all
$n$. (d) Since $(s_n)$ nonincreasing and bounded below by
$1/2$,
$\lim s_n=s_0\in{\Bbb R}$ exists. Using a limit theorem on
$s_{n+1}=(s_n+1)/3$ we have $s_0=\lim s_{n+1}=\lim
(s_n+1)/3=(\lim
s_n+1)/3=(s_0+1)/3\Longleftrightarrow s_0=1/2$.
\n {\bf \textcolor{red}{[\#11.8*]}} (a) Let
$S_N=\{s_n:n>N\}$.
Then by Ex.5.4, $\inf
\{s_n:n>N\}=-\sup(-\{s_n:n>N\})=-\sup\{-s_n:n>N\}$.
Taking limit
in $N\to\infty$, then by definition and a limit theorem we
have
$\liminf
s_n=\lim_{N\to\infty}\inf\{s_n:n>N\}=\lim_{N\to\infty}(-\sup\{-s_n:n>N\})
=-\lim_{N\to\infty}\sup\{-s_n:n>N\}=-\limsup (-s_n)$. (b)
Obviously $(-t_k)$ is monotone iff $(t_k)$ is monotone. Then
by a
limit theorem and (a) we have
$\lim_{k\to\infty}(-t_k)=-\lim_{k\to\infty}t_k=-(\limsup
(-s_n))=\liminf s_n$.
\n {\bf \textcolor{red}{[\#11.10*]}} (a) $S=\{1/n:n\in{\Bbb
N}\}\cup\{0\}$. In fact, the $n$th column subsequence
converges to
$1/n$, and every row subsequence converges to 0. So
$S\supset\{1/n:n\in{\Bbb N}\}\cup\{0\}$. Moreover, for any
number
$a\notin S$, there is a small $\epsilon_0>0$ such that
the
interval $(a-\epsilon_0,a+\epsilon_0)$ contains no points of
the
sequence $(s_n)$, and therefore $a$ cannot be a
subsequential
limit of $(s_n)$, and $S=\{1/n:n\in{\Bbb N}\}\cup\{0\}$
follows.
(b) By inspection, $\limsup s_n=1=\sup S,\liminf s_n=0=\inf
S$.
\n {\bf \textcolor{red}{[\#12.2]}} Since $0\le \liminf
|s_n|\le\limsup |s_n|$ for any sequence, then
$\limsup|s_n|=0$ iff
$\liminf |s_n|=\limsup |s_n|=0$ iff $\lim|s_n|=0$ iff $\lim
s_n=0$.
\n {\bf \textcolor{red}{[\#12.4]}} Follow the hint.
\n {\bf \textcolor{red}{[\#12.6]}} Follow the hint.
\n {\bf \textcolor{red}{[\#12.8*]}} Because every sequence
has a
subsequence converging to its limsup (Rmk: state known
results
rather than theorem, corollary, or lemma numbers from text,
such
as Corollary 11.4 in this case. Follow this convention when
you
take exams), there is a subsequence $r_{n_k}=s_{n_k}t_{n_k}$
of
the product sequence $r_n:=s_nt_n$ such that
$\lim_{k\to\infty}s_{n_k}t_{n_k}=\limsup s_nt_n$. (This does
not
imply $s_{n_k}$ or $t_{n_k}$ convergent!). Because every
bounded
sequence has a converging subsequence, and $(s_n),(t_n)$,
hence
$(s_{n_k}), (t_{n_k})$ automatically, are bounded,
$(s_{n_k})$ has
a converging subsequence $(s_{n_{k_l}})$ to $s\in{\Bbb R}$
(This
does not imply $(t_{n_{k_l}})$ converges). By the same
result,
$(t_{n_{k_l}})$ has a converging subsequence
$(t_{n_{k_{l_m}}})$
to $t\in{\Bbb R}$. Now we have found convergent subsequences
$(t_{n_{k_{l_m}}})$, $(s_{n_{k_{l_m}}})$. Because all
subsequences
of a convergent sequence converge to the same limit, we have
\[
\lim_{m\to\infty}s_{n_{k_{l_m}}}t_{n_{k_{l_m}}}=\lim_{k\to\infty}s_{n_k}t_{n_k}=\limsup
s_nt_n.
\]
On the other hand, by the limit product theorem we have
\begin{equation*}
\begin{split}
\lim_{m\to\infty}s_{n_{k_{l_m}}}t_{n_{k_{l_m}}}&=\lim_{m\to\infty}s_{n_{k_{l_m}}}
\lim_{m\to\infty}t_{n_{k_{l_m}}}\\
&\le \limsup s_n\limsup t_n.
\end{split}
\end{equation*}
The last inequality holds because $\limsup$ of every
sequence is
the least upper bound of all the sequential limits of the
sequence, and the fact that both $s_n, t_n$ are nonnegative.
\n {\bf \textcolor{blue}{A simpler, alternative proof by
Kirsty:}}
For any $n>N$ we have $S_N=\sup\{s_n:n>N\}\ge s_n\ge
0,T_N=\sup\{t_n:n>N\}\ge t_n\ge 0$ $\Longrightarrow
s_nt_n\le
S_NT_N\Longrightarrow \sup\{s_nt_n:n>N\}\le S_NT_N$. By
definition
of $\limsup$ and the product limit theorem, we have
\begin{equation*}
\begin{split}
\limsup s_nt_n&=\lim_{N\to\infty}\{s_nt_n:n>N\}\\
&\le\lim_{N\to\infty}S_NT_N=\lim_{N\to\infty}S_N\lim_{N\to\infty}T_N\\
&=\limsup s_n\limsup t_n.\qquad\qquad \square
\end{split}
\end{equation*}
\n {\bf \textcolor{red}{[\#12.12*]}} Following the hint, we
have
for any $n>M>N$,
\begin{equation}\label{3rdinequality}
\begin{split}
\sigma_n&=\frac{s_1+\cdots +s_n}{n}=\frac{s_1+\cdots
+s_N}{n}+\frac{s_{N+1}+\cdots +s_n}{n}\\
&\le\frac{s_1+\cdots
+s_N}{M}+\frac{n-N}{n}\sup\{s_n:n>N\}\\
&\hbox{(for $s_n\ge 0, n>M$)}\\
&<\frac{s_1+\cdots +s_N}{M}+\sup\{s_n:n>N\}\\
&\hbox{(for $\frac{n-N}{n}<1,n>N$ and $s_n\ge
0$)}\\
\end{split}
\end{equation}
Since it holds for all $n>M$, it holds for
$\sup\{\sigma_n:n>M\}$
\[
\sup\{\sigma_n:n>M\}\le \frac{s_1+\cdots
s_N}{M}+\sup\{s_n:n>N\}
\]
By definition and the fact that limits preserve inequality
relations we have $\limsup
\sigma_n=\lim_{M\to\infty}\sup\{\sigma_n:n>M\}\le
\lim_{M\to\infty}[\frac{s_1+\cdots
s_N}{M}+\sup\{s_n:n>N\}]=\sup\{s_n:n>N\}$. Since this
inequality
holds for all $N$,$\limsup
\sigma_n\le\lim_{N\to\infty}\sup\{s_n:n>N\}=\limsup s_n$
follows.
To show $\liminf s_n\le\liminf \sigma_n$, we argue similarly
as in
(\ref{3rdinequality}) above as follows:
\begin{equation*}
\begin{split}
\sigma_n&=\frac{s_1+\cdots s_n}{n}=\frac{s_1+\cdots
+s_N}{n}+\frac{s_{N+1}+\cdots +s_n}{n}\\
&\ge\frac{n-N}{n}\inf\{s_n:n>N\}\hbox{(for $s_n\ge
0$)}\\
&=1-\frac{N}{n}\inf\{s_n:n>N\}\\
&>1-\frac{N}{M}\inf\{s_n:n>N\}\hbox{
(for $s_n\ge 0,n>M$)}\\
\end{split}
\end{equation*}
Taking the limits in the order of $M\to\infty$ first and
$N\to\infty$ afterwards gives rise to the required result.
\n {\bf \textcolor{red}{[\#14.4]}} (a) Use Comparison Test.
$1/[n+(-1)^n]^2\le 1/(n-1)^2$ and $\sum_{n=2}^\infty
1/(n-1)^2=\sum_{n=1}^\infty 1/n^2$ converges. (b) The
partial sum
$s_n=(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+\cdots
+(\sqrt{n}-\sqrt{n-1})+(\sqrt{n+1}-\sqrt{n})=\sqrt{n+1}-1\to\infty$.
Diverges. (c) By Ratio Test, $|a_{n+1}/a_n|=1/(1+1/n)^n\to
1/e<1\Longrightarrow $ converges.
\n {\bf \textcolor{red}{[\#14.6]}} Let $B$ be an upper bound
of
$(|b_n|)$: $|b_n|\le B \forall n$. By Cauchy criterion and
the
assumption that $\sum|a_n|<\infty$, we have $\forall
\epsilon>0,
\exists N$ s.t. $\forall m>n>N \Longrightarrow
\sum_{k=n+1}^m|a_k|<\epsilon/B$. Hence
$\sum_{k=n+1}^m|a_kb_k|\le
B\sum_{k=n+1}^m|a_k|<B\epsilon/B=\epsilon$. This proves
by Cauchy
criterion that $\sum a_nb_n$ converges absolutely.
\n {\bf \textcolor{red}{[\#14.8]}} Use this inequality:
$(a+b)^2=a^2+2ab+b^2\ge ab$ and the Comparison Test.
\n {\bf \textcolor{red}{[\#14.12*]}} Since every sequence
has a
converging subsequence to its liminf, we have in this case a
subsequence $a_{n_k}$ such that $|a_{n_k}|\to
\liminf|a_n|=0$.
Thus, w.l.o.g., we assume $|a_n|\to 0$ as $n\to\infty$. We
next
construct a subsequence $a_{n_k}$ such that $|a_{n_k}|\le
\frac{1}{k^2}$ (without specifying, it implies automatically
that
$n_1<n_2<\cdots<n_k<\cdots$.) We do this by
induction using the
assumption that $a_n\to 0$. By definition, for
$\epsilon=1/1^2=1$,
$\exists N$ s.t. $n>N\Longrightarrow
|a_n|=|a_n-0|<\epsilon=1/1^2$. Define $n_1=N+1$. Assume
$a_{n_i},i=1,2,\dots,k$ are found. Then to construct
$a_{n_{k+1}}$
we use again the assumption that $a_n\to 0$. To this end,
let
$\epsilon=1/(k+1)^2$. Then $\exists N$ s.t.
$n>N\Longrightarrow
|a_n|<\epsilon=1/(k+1)^2$. Define
$n_{k+1}=\max\{N+1,n_k+1\}$ then
we have $n_{k+1}>n_k$ and $|a_{n_{k+1}}|<1/(k+1)^2$ as
required.
Hence by induction $(a_{n_k})$ can be constructed with
$|a_{n_k}|<1/k^2$. Since $\sum \frac{1}{k^2}<\infty$
converges, by
the Comparison Test, $\sum_{k=1}^\infty a_{n_k}$ converges
absolutely, and itself converges as well.
\n {\bf \textcolor{red}{[\#14.14*]}} Let $s_n$ be the $n$th
partial sum of this series $\sum
a_n=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\cdots$. Then $s_n$
is a
monotone increasing sequence. Notice that there are exactly
$2^{k-1}$ terms of the form $\frac{1}{2^k}$, all together
there
are $1+2+2^2+\cdots 2^{k-1}=2^k-1$ terms for all the terms
having
the form $\frac{1}{2^i}$ with $i=1,2,\dots, k$. Hence the
$(2^k-1)$st partial sum is
\[
s_{2^k-1}=\sum_{i=1}^k\sum_{j=1}^{2^{i-1}}\frac{1}{2^i}=\sum_{i=1}^k\frac{1}{2}=k/2.
\]
Hence $s_{2^k-1}=k/2\to\infty$, and $s_n\to\infty$ follows.
It is
obvious that $a_n<\frac{1}{n}$, and then by the
Comparison Test we
conclude that $\sum \frac{1}{n}$ diverges as well.
\n {\bf \textcolor{red}{[\#15.4*]}} (a) Either by
Comparion/Integral or Comparison Test. By
Comparison/Integral
Test, we start off by noticing $\frac{1}{\sqrt{n} \log n}\ge
\frac{1}{n\log n}$. $f(x)=\frac{1}{x\log x}$ is monotone
decreasing for $x\ge 2$. $\sum_{n=2}^\infty\frac{1}{n\log
n}\ge
\int_2^\infty \frac{1}{x\log x}dx=\infty$ because $\int
\frac{1}{x\log x}dx =\log\log x$. Therefore by Integral
Test,
$\sum_{n=2}^\infty\frac{1}{n\log n}$ diverges, and by
Comparison
Test $\frac{1}{\sqrt{n} \log n}\ge \frac{1}{n\log n}
=\infty$
diverges as well. By Comparison Test alone, we notice $\log
n<\sqrt n$ for $n\ge 1$, and $\frac{1}{\sqrt{n}\log n}\ge
\frac{1}{n}$. Since $\sum \frac{1}{n}$ diverges, $\sum
\frac{1}{\sqrt{n}\log n}$ diverges. (b) Either by Comparison
or
Integral Test. Use Comparison Test we have $\frac{\log
n}{n}\ge
\frac{1}{n}$ for $n\ge 3$ (assuming $\log $ the natural
logarithmic, or $n> 10$ if the base 10 logarithmic) and
the
divergence follows from the divergence of $\sum
\frac{1}{n}$. Use
Integral Test, we check first that $f(x)=\frac{\log x}{x}$
is
monotone decreasing which is the case for $x\ge 3$ since
$f'(x)=\frac{1-\log x}{x^2}<0$. Because
$\int_3^\infty\frac{\log
x}{x}dx=\frac{(\log x)^2}{2}|_3^\infty =\infty$, the series
$\sum
\frac{\log n}{n}$ diverges as well. (c) Use Integral Test on
$f(x)=\frac{1}{x\log x(\log \log x)}$. (d) Use Integral Test
or
Comparison Test. By Integral Test, we use $f(x)=\frac{\log
x}{x^2}$ which is monotone decreasing since
$f'(x)=\frac{1-2\log
x}{x^3}<0$ for $x\ge 2$. Also $\int\frac{\log
x}{x^2}dx=-\frac{\log x}{x}+\int \frac{1}{x^2}dx=-\frac{\log
x}{x}-\frac{1}{x}$ using integration by parts. Hence
$\int_2^\infty\frac{\log x}{x^2}dx=\frac{1+\log 2}{2}$
converges.
By Integral Test, the series converges. Use Comparison Test,
we
note that $\log n<n^q$ for any fixed $0<q<1$ and
sufficiently
large $n>N$. So $\frac{\log
n}{n^2}<\frac{n^q}{n^2}=\frac{1}{n^p}$
with $p=2-q>1$. Since $\sum \frac{1}{n^p}$ converges for
any
$p>1$, we conclude by the Comparison Test that $\sum
\frac{\log
n}{n^2}<\sum \frac{1}{n^p}<\infty$ converges.
\n {\bf \textcolor{red}{[\#15.6*]}} (a) $a_n=1/n$. (b) $\sum
a_n<+\infty, a_n\ge 0$ implies $a_n^2\le a_n$ for all
large $n$
since $a_n\to 0$. Comparison Test. (c)
$a_n=(-1)^n1/\sqrt{n}$.
\n {\bf \textcolor{red}{[\#17.13*]}} (a) For any $x\in{\Bbb
R}$
and $n\in{\Bbb N}$, there is a rational $r_n\in{\Bbb Q}$
such that
$x<r_n<x+1/n$ by the Archimedean Property. Hence the
rational
number sequence $r_n\to x$. Also the irrational number
sequence
$t_n=r_n+\sqrt{2}/n\to x$. Therefore either $\lim
f(r_n)=1\ne
f(x)$ if $x\in{\Bbb R}-{\Bbb Q}$ or $\lim f(t_n)=0\ne f(x)$
if
$x\in{\Bbb Q}$. $f$ is not continuous in both cases. (b)
Similar
to (a) when $x\ne 0$. For $x=0$, we always have
$|h(x)-h(0)|=|h(x)|\le |x|$ to which $\epsilon-\delta$
argument
can be easily fashioned.
\n {\bf \textcolor{red}{[\#17.14*]}} If $x\in{\Bbb Q}$,
construct
an irrational sequence $t_n\in{\Bbb R}-{\Bbb Q}$ in the same
way
as in \#17.13 above so that $t_n\to x$ and $\lim f(t_n)=0\ne
f(x)=1/q$ if $x\ne 0$. If $x\in{\Bbb R}-{\Bbb Q}$, $f(x)=0$
and we
claim for every sequence $x_n\to x$, $\lim f(x_n)\to 0$.
Otherwise, there is a sequence $x_n\to 0$ but
$f(x_n)\nrightarrow
0$, and w.l.o.g we assume $|f(x_n)|\ge \epsilon_0>0$ for
some
constant $\epsilon_0$. This implies then that $x_n\in{\Bbb
Q}$ and
$f(x_n)=1/q_n\ge \epsilon_0$, which in turn implies
$0<q_n\le
A=1/\epsilon_0$. Since convergent sequences are bounded,
$|p_n/q_n|=|x_n|\le B$ for some constant $B$. Hence
$|p_n|\le
B|q_n|\le BA$. Therefore there are only finitely many
parings of
$p_n,q_n$ for $|p_n|\le BA, 0<q_n\le A$. Therefore the
sequence
$x_n$ can only take on finitely many values. Since sequence
$(x_n)$ converges, $x_n$ must take on a fixed number for all
large
$n$ and that fixed number is on of the rationals: $p_n/q_n$
with
$|p_n|\le BA$ and $0<q_n\le A$. This contradicts to the
fact that
$x_n$ converges to an irrational number.
\n {\bf \textcolor{red}{[\#17.17*]}} It is obvious that the
condition is necessary since it is a special case of the
definition that $x_n\to x_0$ implies $f(x_n)\to f(x_0)$.
Conversely, assume the contrary that there is a sequence
$x_n\to
x_0$ with $x_n\in{\rm dom}(f)$ but $\lim f(x_n)\nrightarrow
f(x_0)$. Then $\exists \epsilon_0>0$ so that $\forall N,
\exists
n\ge N$ with $|f(x_n)-f(x_0)|\ge \epsilon_0$. That is a
subsequence can be found so that $|f(x_n)-f(x_0)|\ge
\epsilon_0$.
Therefore w.o.l.g, we assume $x_n$ is such a subsequence.
Then we
concluded right away that $x_n\ne x_0$ but $x_n\to x_0$
nonetheless. This contradicts the assumption that we must
have
$f(x_n)\to f(x_0)$ whenever $x_n\to x_0$ and $x_n\ne x_0$
for all
$n$.
\n {\bf \textcolor{red}{[\#18.4*]}} $f(x)=1/(x-x_0)$.
\n {\bf \textcolor{red}{[\#18.5*]}} (a) Let $h=f-g$. Then
$h$ is
continuous as both $f$ and $g$ are continuous. Also
$h(a)=f(a)-g(a)\le 0$ and $h(b)=f(b)-g(b)\ge 0$ by
assumption.
Then by the Intermediate Value Theorem $h(x_0)=0$ for some
$x_0\in[a,b]$, implying $f(x_0)=g(x_0)$ as required. (b) Let
$g(x)=x$. Then $f(0)\ge 0=g(0)$ and $f(1)\le 1=g(1)$ by the
assumption that $f$ maps $[0,1]$ into $[0,1]$. Hence the
conditions of (a) are satisfied for the given functions
$f,g$ and
$[a,b]=[0,1]$.
\n {\bf \textcolor{red}{[\#18.10*]}} Let $g(x)=f(x+1)-f(x),
x\in
[0,1]$. Then $g$ is continuous in $[0,1]$ as $f$ is
continuous in
$[0,2]$. Also, $g(0)=f(1)-f(0), g(1)=f(2)-f(1)=f(0)-f(1)$ by
the
assumption that $f(2)=f(0)$. Therefore
$g(0)=-(f(1)-f(0))=-g(1)$,
implying either $g(0)=g(1)=0$ or $g(0)$ and $g(1)$ have
opposite
signs. In the latter case there exists a number $x_0\in [0,1]$
such that $g(x_0)=0$ by IVT. In either cases the same result
holds. Therefore with $y_0=x_0+1$ $f(y_0)=f(x_0)$ follows.
\n {\bf \textcolor{red}{[\#19.2*]}} (c) Only. $\forall
\epsilon>0$, let $\delta =\epsilon/4$ s.t.
$|x-y|<\delta, x,y\ge
1/2$ implies
$|f(x)-f(y)|=|\frac{1}{x}-\frac{1}{y}|=|\frac{y-x}{xy}|\le
|\frac{y-x}{(1/2)(1/2)}|=4|x-y|<4\delta=\epsilon$.
\n {\bf \textcolor{red}{[\#19.7*]}} (a) Note first that the
continuity of $f$ on $[0,\infty)$ implies the continuity of
$f$ on
any subset, including $[0,k+1]$. Since $[0,k+1]$ is bounded
and
closed interval of ${\Bbb R}$ $f$ is uniformly continuous on
$[0,k+1]$. We now show that $f$ is uniformly continuous on
$[0,\infty)$ by definition. $\forall \epsilon>0$,
$\exists
\delta_1>0$ s.t. $|x-y|<\delta_1,x,y\in [k,\infty)$
implies
$|f(x)-f(y)|<\epsilon$ by the assumption that $f$ is
uniformly
continuous on $[k,\infty)$. Since $f$ is uniformly
continuous on
$[0,k+1]$, $\exists \delta_2>0$ s.t. $|x-y|<\delta_2,
x,y\in
[0,k+1]$ implies $|f(x)-f(y)|<\epsilon$. Let
$\delta=\min\{1,\delta_1,\delta_2\}$, we claim that
$|x-y|<\delta,
x,y\in[0,\infty)$ implies $|f(x)-f(y)|<\epsilon$. To this
end all
we need to show is that the condition $|x-y|<\delta,
x,y\in[0,\infty)$ implies either $x,y\in[0,k+1]$ or $x,y\in
[k,\infty)$. Suppose $x,y\in[0,k+1]$ does not hold. Then
either
both $x,y\ge k+1>k$ or one of $x,y$ is in $[0,k+1]$ while
the
other is not. In the former case we have $x,y\in
[k,\infty)$. In
the latter case, suppose w.o.l.g that $x<k+1\le y$. Then
the
condition that $|x-y|<\delta\le 1$ implies that
$x=y-(y-x)\ge
y-|y-x|\ge y-\delta> k+1-1=k$. That is, $k<x<y$ and
$x,y\in
[k,\infty)$ holds as well. (b) Obviously $f(x)=\sqrt x$ is
continuous in $[0,\infty)$. Because
$f'(x)=\frac{1}{2}\frac{1}{\sqrt x}\le
\frac{1}{2}\frac{1}{1}$ for
$x\ge 1$, hence $f$ is uniformly continuous on $[1,\infty)$.
By
(a) $f$ is uniformly continuous in $[0,\infty)$.
\n {\bf \textcolor{red}{[\#19.9*]}} (a)
$f(x)=x\sin(\frac{1}{x})$
for $x\ne 0$ and $f(0)=0$ is continuous on ${\Bbb R}$. This
follows from the product and composition rules for
continuous
functions when $x\ne 0$ and the estimate $|f(x)-f(0)|\le
|x|$ at
the point $0$. (b) Let $S$ be any bounded subset of ${\Bbb
R}$.
Then $a=\inf S$ and $b=\sup S$ are all finite numbers.
Therefore
$f$ is uniformly continuous in the bounded, closed interval
$[a,b]$ since $f$ is continuous there. Hence $f$ is
uniformly
continuous on any subset of $[a,b]$ which includes $S$. (c)
Because
$f'(x)=\sin(\frac{1}{x})-\frac{1}{x}\cos(\frac{1}{x})$ for
$x\ne 0$ we have $|f'(x)|\le 1+1 =2 $ for $|x|\ge 1$. Hence
$f$ is
uniformly continuous in $(-\infty, -1]$ and $[1,\infty)$.
Because
$f$ is also uniformly continuous in, say $[-2,2]$, the exactly
same argument for \#19.7(a) can be used to show $f$ is
uniformly
continuous in ${\Bbb R}$.
\n {\bf \textcolor{blue}{[Notes Supplement On Riemann
Integral]}}
Let $U(\{x_i\})=\sum \sup_{I_i}f\Delta x_i$ be the upper sum
of
any partition $a=x_0<x_1<\dots<x_n=b, \Delta
x_i=x_i-x_{i-1},
i=1,2,\dots, n, I_i=[x_{i-1},x_i]$. We claim that
$\lim_{\Delta
x\to 0}U(\{x_i\})=\ell$ exists where $\Delta x =\max\{\Delta
x_i,i=1,2,\dots, n\}$.
Since $f$ is continuous in $[a,b]$, $\exists \bar x_i\in
[x_{i-1},x_i]$ such that $f(\bar x_i)=\sup_{I_i}f$. Since
$f$ is
uniformly continuous in $[a,b]$, then $\forall
\epsilon>0$,
$\exists \delta>0$ s.t. $|x-y|<\delta, x,y\in [a,b]$
implies
$|f(x)-f(y)|<\epsilon$.
We now proceed to prove the claim by first developing a background
result. A partition $a=y_0<y_1<\dots y_m=b$ is said to
be a {\em
refinement} of a given partition
$a=x_0<x_1<\dots<x_n=b$ if
$\{x_i\}$ is just a subset of $\{y_j\}$, i.e., $x_i=y_{j_i}$
for
some $j_i$ and for all $i=0,1,2,\dots, n$. Then the difference
between the corresponding upper sums $U(\{x_i\})-U(\{y_j\})$
has
the following properties.
Either a subinterval $[x_{i-1},x_i]$ contains no refinement
points
$y_j$ with $x_{i}=y_{j_i}$ for some $j_i$ and
$x_{i-1}=y_{j_i-1}$.
In this case the corresponding summands $\sup_{I_i}f\Delta
x_i$
and $\sup_{J_{j_i}}f\Delta y_{j_i}$ are identical and cancel
out
each other in the difference $U(\{x_i\})-U(\{y_j\})$.
Or a subinterval $[x_{i-1},x_i]$ contains some refinement
points
$x_{i-1}=y_j<y_{j+1}<\cdots<y_{j+k}=x_i$ for some
$k>1$. In this
case the summand $\sup_{I_i}f\Delta x_i$ corresponds to the
subsum
$\sum_{l=1}^k\sup_{J_{j+l}}f\Delta y_{j+l}$ with
$J_{j+l}=[y_{j+l-1},y_{j+l}]$. Breaking up $[x_{i-1},x_i]$
according to its refinement
$x_{i-1}=y_j<y_{j+1}<\cdots<y_{j+k}=x_i$, the
corresponding
difference in absolute value $|\sup_{I_i}f\Delta
x_i-\sum_{l=1}^k\sup_{J_{j+l}}f\Delta y_{j+l}|$ becomes
\[
|\sum_{l=1}^k(\sup_{I_i}f-\sup_{J_{j+l}}f)\Delta
y_{j+l}|<\sum_{l=1}^k\epsilon\Delta y_{j+l}\le \epsilon
\Delta x_i
\]
if $\Delta x_i=x_i-x_{i-1}<\delta$ by the uniformly
continuity
since $\sup_{I_i}f=f(\bar x_i)$ and $\sup_{J_{j+l}}f=f(\bar
y_{j+l})$ with $\bar x_i,\bar y_{j+l}\in
[x_{i-1},x_i]\Longrightarrow |\bar x_i-\bar y_{j+l}|\le
x_i-x_{i-1}<\delta$. Hence the upper sum difference in
absolute
value $|U(\{x_i\})-U(\{y_j\})|$ on a whole is bounded above
by
$\epsilon\sum_{i=1}^n\Delta x_i=\epsilon (b-a)$ for any
refinement
of partition $\{x_i\}$ and $\Delta x\le \delta$.
We are now ready to prove the claim $\lim_{\Delta x\to
0}U(\{x_i\})=\ell$. As we did in class we first show that
the
sequence $U_n=U(\{x_i\})$ in regular partition
$x_i=a+i\Delta
x,\Delta x=(b-a)/n$ has a limit. We do this by showing that
$\{U_n\}$ is a Cauchy sequence. In fact, for
$(b-a)/N<\delta$ or
$N>(b-a)/\delta$ and any $m,n>N$, the partition for
$U_{mn}$ is a
refinement for both partitions of $U_n$ and $U_m$ because
the
partition points of $U_n$ satisfy
$x_i=a+i\frac{b-a}{n}=a+im\frac{b-a}{nm}$ which is a
partition
point of $U_{mn}$ for each $i$ and similarly for $U_m$. By
what we
have just proved above,
$|U_n-U_m|=|U_n-U_{mn}+U_{mn}-U_m|\le
|U_n-U_{mn}|+|U_{mn}-U_m|<2\epsilon (b-a)$ since $\Delta
x=(b-a)/n$ for $U_n$ and $\Delta x=(b-a)/m$ for $U_m$ are
both
less than $(b-a)/N<\delta$ for $m,n>N$. This shows
$U_n$ is Cauchy
and $\lim U_n=\ell$ follows.
Finally we prove $\lim_{\Delta x\to 0}U(\{x_i\})=\ell$ for
all
partition. Assume on the contrary that it is false, then a
sequence of upper Riemann sums $U(\{x_i^k\}), k=1,2,\dots$
can be
found such that $|U(\{x_i^k\})-\ell|>\epsilon_0$ for some
fixed
number $\epsilon_0$ even though $\Delta x^k=\max\{\Delta
x_i^k,
i=1,2,\dots n_k\}\to 0$ as $k\to 0$. Because $U_n\to \ell$,
we
have $N_0>0$ such $n>N_0$ implies
$|U_n-\ell|<\epsilon_0/2$ and
thus $|U(\{x_i^k\})-U_n|=|U(\{x_i^k\})-\ell-(U_n-\ell)|\ge
|U(\{x_i^k\})-\ell|-|U_n-\ell|
>\epsilon_0/2$ for $n>N_0$ and all $k$. This has to be
a
contradiction for the following reasons. For each regular
$n$th
partition and any given one in $\{x_i^k\}$, putting all
these
points together to form a refinement partition for both
$U_n$ and
$U(\{x_i^k\})$ and denote the refinement upper sum by
$U_n^k$.
Then when $\Delta x^k,(b-a)/n<\delta$, we have
$|U(\{x_i^k\})-U_n|=|U(\{x_i^k\})-U_n^k+U_n^k-U_n|\le
|U(\{x_i^k\})-U_n^k|+|U_n^kU_n|<2\epsilon
(b-a)<\epsilon_0/2$
since $\epsilon$ is arbitrary. This completes the proof.
\end{document}