\documentclass[12pt]{amsart}

% Package Being Used:

\usepackage{enumitem}
\usepackage{bigints}
\usepackage{amssymb}
\usepackage{bm}
\usepackage{graphicx}
\usepackage{psfrag}
\usepackage{color}
\usepackage{url}
\usepackage{algpseudocode}
\usepackage{fancyhdr}
\usepackage{xy}
\input xy
\xyoption{all}

%\usepackage{soul} % this is a package used only temporary that allows us to underline comments that we are planning to remove in future versions of the paper.

% Paper Format and Geometry:

\voffset=-1.4mm
\oddsidemargin=17pt
\evensidemargin=17pt
\topmargin=26pt
\headheight=9pt     
\textheight=576pt
\textwidth=440.8pt
\parskip=0pt plus 4pt

% Head Labels:

\pagestyle{fancy}
\fancyhf{}
\renewcommand{\headrulewidth}{0pt}
\renewcommand{\footrulewidth}{0pt}

% Theorems-like Format and Numbering:


\newtheorem{theorem}{Theorem}%[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{cor}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\numberwithin{equation}{section}

% Personalized Commands:

\newcommand\cc{\mathbb{C}}
\newcommand\ff{\mathbb{F}}
\newcommand\nn{\mathbb{N}}
\newcommand\qq{\mathbb{Q}}
\newcommand\rr{\mathbb{R}}
\newcommand\zz{\mathbb{Z}}

\newcommand\pval{\mathsf{v}_p}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
	
	\mbox{}
	\title{Mock Quiz 2}
	
	\maketitle
	
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%\section{Introduction} \label{sec:intro}
	\noindent {\footnotesize Note: Most of the problems were taken from the textbook} \cite{jS15}.
		
	\begin{problem}
		If $a,b \in \rr$ such that $a,b > 0$, compute the limit of the sequence $\{a_n\}$, where
		\[
			a_n = \bigg(1 + \frac an \bigg)^{\frac nb}.
		\]
	\end{problem}
	
	\noindent {\it Solution 1:} We can compute the limit of $\{a_n\}$ as follows:
	\begin{align*}
		\lim_{n \to \infty} a_n 
			&= \lim_{n \to \infty} \exp\bigg(\frac nb \ln \bigg(1 + \frac an\bigg)\bigg) = \exp\bigg(\lim_{n \to \infty}\frac nb \ln \bigg(1 + \frac an\bigg)\bigg) \\
			&= \exp \bigg(\lim_{x \to \infty} \frac{\ln\big(\frac {x+a}x\big)}{\frac bx} \bigg) = \exp\bigg(\lim_{x \to \infty}\frac{\frac x{x+a}\frac{-a}{x^2}}{\frac{-b}{x^2}}\bigg) \\
			&= \exp\bigg(\lim_{x \to \infty} \frac{a}{b} \frac{x}{x+a}\bigg) = e^{\frac ab}.
	\end{align*}
	
	\noindent {\it Solution 2:} We can compute the limit of $\{a_n\}$ as follows:
	\begin{align*}
		\lim_{n \to \infty} a_n
			&= \lim_{n \to \infty} \bigg(1 + \frac an \bigg)^{\frac nb} = \lim_{n \to \infty} \bigg(\bigg(1 + \frac an\bigg)^{\frac na}\bigg)^{\frac ab} = \bigg( \lim_{n \to \infty} \bigg(1 + \frac an\bigg)^{\frac na}\bigg)^{\frac ab} = e^{\frac ab}.
	\end{align*}
	
	\hfill \qed
	
	\begin{problem}
		Decide whether the series
		\begin{equation} \label{eq:problem 2}
			\sum_{n=2}^\infty \ln\bigg(\frac{n+1}{n}\bigg)
		\end{equation}
		converges. If it is convergent, then find the limit.
	\end{problem}
	
	\noindent {\it Solution:} Denoting the $n$-th partial sum of \eqref{eq:problem 2} by $s_n$, we can observe that
	\[
		s_n = \sum_{k=2}^n \ln\bigg(\frac {k+1}k\bigg) = \sum_{k=2}^n \big(\ln(k+1) - \ln k\big) = \ln (n+1) - \ln 2.
	\]
	Therefore
	\[
		\sum_{n=2}^\infty \ln\bigg(\frac{n+1}{n}\bigg) = \lim_{n \to \infty} \ln (n+1) - \ln 2 = \infty.
	\]
	\hfill \qed
	
	\begin{thebibliography}{20}
		
		\bibitem{jS15} J.~Stewart: \emph{Single Variable Calculus} 8th Edition, Cengage Learning, Boston 2015.
		
	\end{thebibliography}
	
\end{document}