\documentclass[12pt,a4paper]{amsart}
%%%%%%%%%%% Here is where you tell LaTeX to use packages. Each
%%%%%%%%%%% package gives you access to different commands and
%%%%%%%%%%% functionality. Most of them are unnecessary for the
%%%%%%%%%%% mathematics in this document.

%%%%%%%%%%% If you hadn't figured out, the % symbol automatically
%%%%%%%%%%% nulls out all subsequent characters on that line. So
%%%%%%%%%%% it's used for commenting code.
\usepackage{hyperref}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{framed}
\usepackage{multicol}
\usepackage[OT2,T1]{fontenc}
\usepackage{graphicx}
\usepackage[all]{xy}
\usepackage{color}
\usepackage{mathrsfs}
\usepackage{mathtools}

%%%%%%%%%%% Determining margins and spacing.
\setlength\topmargin{-0.3in}
\addtolength{\oddsidemargin}{-.625in}
	\addtolength{\evensidemargin}{-.625in}
	\addtolength{\textwidth}{1.25in}
\setlength\footskip{1in}

%%%%%%%%%%% I think this command enables `sections'
\numberwithin{equation}{section}

%%%%%%%%%%% This command allows you to label things as `theorem',
%%%%%%%%%%% `conjecture', etc.
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{question}{Question}[theorem]
\newtheorem{example}{Example}[theorem]


\usepackage[utf8]{inputenc}
\title{Math 220.201 Class 25 Questions}
\begin{document}
\maketitle
\begin{enumerate}

\item [1.] Prove that the sequence $\left\{ \frac{4n^2-3}{5n^2-2n}\right\}$ converges to $\frac{4}{5}$.
\vskip 0.2in
\begin{proof} [Proof 1]
We can calculate
$$\frac{4n^2-3}{5n^2-2n}-\frac{4}{5}=\frac{5(4n^2-3)-4(5n^2-2n)}{5(5n^2-2n)}=\frac{8n-15}{25n^2-10n}$$
We must show that for any $\epsilon$, there is some $N$ such that for every $n>N$, this fraction has absolutely value less than $\epsilon$. But $|8n-15|\le |8n|$, and as long as $n\ge 10$, we have $|25n^2-10n|\ge |24n^2|$. Therefore, as long as $n\ge 10$,
$$\left |\frac{8n-15}{25n^2-10n}\right |\le \left |\frac{8n}{24n^2}\right|=\left|\frac{1}{3n}\right|$$
Therefore, if one takes $n> \mathrm{max}(10, 1/3\epsilon)$, we will have $|\frac{4n^2-3}{5n^2-2n}-\frac{4}{5}|<\epsilon$, as desired.
\end{proof}
\begin{proof} [Proof 2]
Use the limit laws.
$$\lim\limits_{n\to \infty}\frac{4n^2-3}{5n^2-2n}=\lim\limits_{n\to \infty}\frac{4-\frac{3}{n^2}}{5-\frac{2}{n}}=\frac{4-\lim\limits_{n\to \infty}\frac{3}{n^2}}{5-\lim\limits_{n\to\infty}\frac{2}{n}}=\frac{4}{5}$$
\end{proof}
\vskip 0.2in
\item [2.] (12.11) Prove (using the definition of convergence) that if a sequence $\{s_n\}$ converges to $L$, then the sequence $\{s_{n^2}\}$ converges to $L$.
\vskip 0.2in
\begin{proof}
$\{s_n\}$ converges to $L$, which means that for every $\epsilon>0$, there exists some $N$ such that if $n>N$, then $|s_n-L|<\epsilon$. If $n>N$, then $n^2>N$ as well, because $n^2\ge n$ for every natural number $n$. It therefore follows that if $n>N$, $|s_{n^2}-L|<\epsilon$. Thus, for every $\epsilon$, the same value of $N$ works for the sequence $\{s_{n^2}\}$ as for the sequence $\{s_n\}$.
\end{proof}
\vskip 0.2in
\item [3.] (Adapted from 2011 WT2 Q7) Let $\{b_n\}$ be a sequence defined by
\begin{center}
$b_1=2$ and $b_{n+1}=\frac{b_n+\sqrt{b_n}}{2}$
\end{center}
\begin{enumerate}
\item Prove that $1<b_{n+1}<b_n$ for every $n\in\mathbb{N}$.
\item Prove that $\{b_n\}$ converges to $1$.
\end{enumerate}
\vfill
\vskip 0.2in
\begin{proof}
(a) We prove that $1<b_{n+1}<b_n$ by induction on $n$. The base case, $n=1$, holds true because $b_2=1+\frac{1}{\sqrt{2}}$, which is greater than $1$ and smaller than $2=1+1$.

Let us suppose that $1<b_{n+1}<b_n$. We will show that $1<b_{n+2}<b_{n+1}$. Since $1<b_{n+1}$, it follows that
$$\begin{aligned}
1<\sqrt{b_{n+1}}<b_{n+1} &\implies 2<b_{n+1}+\sqrt{b_{n+1}}<2b_{n+1}\\
&\implies 1<\frac{b_{n+1}+\sqrt{b_{n+1}}}{2}<b_{n+1}\\
&\implies 1<b_{n+2}<b_{n+1}
\end{aligned}$$
which completes the induction.

(b) The sequence $\{b_n\}$ is decreasing and is bounded below by $1$, and therefore it must converge to a limit $L\ge 1$. Let us suppose for a contradiction that $L>1$. Then for any $\epsilon>0$, there exists some $N$ such that if $n>N$, $b_n<L+\epsilon$. Then this implies that
$$\begin{aligned}
b_{n+1}=\frac{b_n+\sqrt{b_n}}{2}&<\frac{L+\epsilon+\sqrt{L+\epsilon}}{2}\\
&= \frac{L+\epsilon+\sqrt{L(1+\frac{\epsilon}{L})}}{2}\\
&< \frac{L+\epsilon+\sqrt{L(1+\frac{\epsilon}{L}+\frac{\epsilon^2}{4L^2})}}{2}\\
&= \frac{L+\epsilon+\sqrt{L(1+\frac{\epsilon}{2L})^2}}{2}\\
&= \frac{L+\epsilon+(1+\frac{\epsilon}{2L})\sqrt{L}}{2}\\
&= \frac{L+\sqrt{L}}{2}+\frac{\epsilon}{2}\left(1+\frac{1}{\sqrt{L}}\right)\\
&< \frac{L+\sqrt{L}}{2}+\epsilon
\end{aligned}$$

Pick any positive number $\epsilon<\frac{L-\sqrt{L}}{4}$. Then it will be the case that $\frac{L+\sqrt{L}}{2}+\epsilon<L-\epsilon$, and thus $b_{n+1}<L-\epsilon$. But this contradicts the assumption that $|b_{n+1}-L|<\epsilon$! Therefore, it was impossible that $L>1$.
\end{proof}
\vskip 0.2in
\item [4.] (2013 WT1, Q8) Prove that the function $f:(-1, 1]-\{0\} \rightarrow \mathbb{R}$ given by $f(x)=x-\frac{1}{x}$ is a bijection.
\vskip 0.2in
\begin{proof}
First, we prove the function is injective. Suppose that $x-\frac{1}{x}=y-\frac{1}{y}$ for some numbers $x, y\in (-1, 1]-\{0\}$. Then
$$x-y=\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}$$
$$\implies (x-y)(xy-1)=0$$
It's impossible to have $xy=1$ by absolute value considerations, unless $x=y=1$. Thus, we have $x-y=0$, which implies $x=y$. So the function is injective.

Now we prove it's surjective. Let $r$ be any real number. Then we solve the equation $x-\frac{1}{x}=r$
$$x^2-rx-1=0 \iff x=\frac{r\pm \sqrt{r^2+4}}{2}$$
If $r$ is zero, then the + root is $1$. If $r$ is positive, the - root lies in $(-1, 1]-\{0\}$ by difference of squares:
$$\frac{r-\sqrt{r^2+4}}{2}=\frac{-4}{2(r+\sqrt{r^2+4})}=-\frac{2}{r+\sqrt{r^2+4}}>-\frac{2}{0+\sqrt{0+4}}=-1$$
If $r$ is negative, the + root lies in $(-1, 1]-\{0\}$ by a similar argument.
\end{proof}
\vskip 0.2in
\item [5.] Give an example of
\begin{itemize}
\item A function $f:\mathbb{Z}\rightarrow\mathbb{Z}$ which is injective, but not surjective.
\vskip 0.1in
{\bf Solution:} Try the function $f(n)=2n$.
\vskip 0.1in
\item A function $g:\mathbb{Z}\rightarrow\mathbb{Z}$ which is surjective, but not injective.
\vskip 0.1in
{\bf Solution:} Try the function $g(n)=\lfloor n/2\rfloor$.
\vskip 0.1in
\item Prove that for any such functions $f$ and $g$, $f\circ g$ cannot possibly be bijective.
\vskip 0.1in
{\bf Solution:} Because $g$ is not injective, there are two distinct integers $x$ and $y$ such that $g(x)=g(y)$. Then $f(g(x))=f(g(y))$, so $f\circ g$ is not injective. So it can't possibly be bijective. (You can show $f\circ g$ is not surjective, either.)
\vskip 0.1in
\item Give such examples with the property that $g\circ f$ is bijective.
\vskip 0.1in
{\bf Solution:} The the examples I gave above, $f(n)=2n$ and $g(n)=\lfloor n/2\rfloor$. Then $g(f(n))=n$.

\end{itemize}

\end{enumerate}
\end{document}