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\noindent
\parbox{2.25in}{\bf Math 3220}
\hfill {\Large \bf Homework \#3} \hfill
\parbox{2.25in}{\bf \hfill Due: Mon., Feb. $11^\mathrm{th}$, 2019}

\vspace{0.2in}

%\noindent {\bf\large Name:} \underline{\qquad\sc\large This is my fabulous name! \qquad}
%
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\begin{description}
\item[\#1 Convergence -- By Definition] Prove that the following sequences converge. Use the definition of convergence. Don't use any ``fancy'' theorems.

\begin{enumerate}[(a)]
\item $\displaystyle \left\{ \dfrac{5n}{n+1} \right\}_{n=1}^\infty$

\item $\displaystyle \left\{ \dfrac{2n+1}{n^3+3n-1} \right\}_{n=1}^\infty$
\end{enumerate}

\item[\#2 Convergence Redux] Redo problem \#1. This time show convergence using nothing more than the convergence of $\{ 1/n \}_{n=1}^\infty$ and some sequence arithmetic. 

{\it Example:} $\dfrac{n^2+1}{3n^2-n+5} = \dfrac{n^2}{n^2} \cdot \dfrac{1+1/n^2}{3-1/n+5/n^2} \to \dfrac{1+0}{3-0+5\cdot 0} = \dfrac{1}{3}$ since $1/n \to 0$ implies $1/n^2  \to 0^2=0$ etc.

\item[\#3 Divergence -- Subsequence Style] $\displaystyle\left\{ \dfrac{(-1)^n n^2+2n+1}{3n^2-n+5} \right\}_{n=1}^\infty$ diverges. Show this using subsequences.

\item[\#4 Convergence Abstractly] Let $\{ a_n \}_{n=1}^\infty$ be a sequence which converges to $L$. 

\begin{enumerate}[(a)]

\item Let $b_n = \dfrac{a_n+a_{n+1}}{2}$. Show that $b_n \to L$

\item Suppose that $a_n \leq M$ for all $n \in \mathbb{N}$. Show that $L \leq M$.

\end{enumerate}

\item[\#5 Proceed with Cauchy!] Suppose that $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ are Cauchy sequences. Show that $\{a_n+b_n\}_{n=1}^\infty$ is Cauchy.

\begin{enumerate}[(a)]
\item Do this using the definition of Cauchy sequences.
\item Reprove this using Theorem 2.20 (i.e., a sequence is Cauchy if and only if it converges) and a theorem from Section 2.2. Cite the Theorem.
\end{enumerate}

\item[RESUBMIT] Type up Homework \#2 Problem \#3 and its solution in \LaTeX. 

\end{description}

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