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\title{Math 3100 Homework Assignment 2 - Due Wednesday February 10th}

\author{Your name here}


\date{\today}

\begin{document}

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\begin{exer}
For a sequence of real numbers $\{ a_n \}_{n = 1}^\infty$, and a real number $A \in \mathbb R$ consider the following conditions:

\begin{equation}\tag{P1}
\forall  N \geq 1, \exists \varepsilon > 0 \textrm{ such that } \forall n \geq N \textrm{ we have } | A - a_n | \leq \varepsilon.
\end{equation}
\begin{equation}\tag{P2}
 \exists N \geq 1 \textrm{ such that } \forall \varepsilon > 0, \textrm{ and } n \geq N \textrm{ we have } | A - a_n | \leq \varepsilon.
\end{equation}
\begin{equation}\tag{P3}
\forall \varepsilon > 0, \exists N \geq 1 \textrm{ such that } \forall n \geq N \textrm{ we have } | A - a_n | \leq \varepsilon.
\end{equation}
\begin{equation}\tag{P4}
\forall N \geq 1, \varepsilon > 0, \textrm{ and } n \geq N \textrm{ we have } | A - a_n | \leq \varepsilon.
\end{equation}
\begin{equation}\tag{P5}
\forall \varepsilon > 0 \textrm{ and } N \geq 1, \exists n \geq N \textrm{ such that we have } | A - a_n | \leq \varepsilon. 
\end{equation}

For each condition above give an intuitive description of the class of sequences that satisfy the condition and describe the relationship between those sequences and the number $A$. 
\end{exer}



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\begin{exer}
If $S \subset \mathbb R$ is a set, let $L(S)$ denote the set of accumulation points of $S$. Prove that there either does or does not exist a set $S \subset \mathbb R$ such that $L(L(S)) \not= \emptyset$, but $L ( L ( L (S))) = \emptyset$. 
\end{exer}





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\begin{exer}
Suppose $S \subset \mathbb R$ is an uncountable set. Prove that $S$ has an accumulation point. 
\end{exer}




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\begin{exer}
Given a number $x \in \mathbb R$ we let $\lfloor{x} \rfloor$ denote floor function of $x$, i.e., $\lfloor{x} \rfloor$ is the greatest integer that is less than or equal to $x$. Prove that for all $x \in \mathbb R$ the sequence 
\[
\left\{ \frac{ \lfloor{x} \rfloor + \lfloor{3x} \rfloor + \cdots + \lfloor{(2n + 1)x} \rfloor }{n^2} \right\}_{n = 1}^\infty
\] 
converges to $x$. 
\end{exer}





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\begin{exer}
Given a sequence $\{ a_n \}_{n = 1}^\infty$ we may define a new sequence $\{ \alpha_n \}_{n = 1}^\infty$ by $\alpha_n = \frac{a_1 + a_2 + \cdots + a_n}{n}$. Show that if $\{ a_n \}_{n = 1}^\infty$ converges to $A \in \mathbb R$ then we also have that $\{ \alpha_n \}_{n = 1}^\infty$ converges to $A$. Give an example where $\{ a_n \}_{n =1 }^\infty$ is divergent but $\{ \alpha_n \}_{n = 1}^\infty$ is convergent.  
\end{exer}




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\begin{exer}
Prove that if $\{ a_n \}_{n = 1}^\infty$ converges to $A$, with $A \geq 0$ and $a_n \geq 0$ for all $n \geq 1$, then the sequence $\{ \sqrt{ a_n } \}_{n = 1}^\infty$ converges to $\sqrt{A}$. 
\end{exer}





%\begin{thebibliography}{9}



%\bibitem{halmosmath}
%Paul R. Halmos
%How to write mathematics.
%\url{https://sites.math.washington.edu/~lind/Resources/Halmos.pdf}

%\bibitem{knuthmath}
%Donald E. Knuth, Tracy Larrabee and Paul M. Roberts
%Mathematical Writing.
%\url{https://jmlr.csail.mit.edu/reviewing-papers/knuth_mathematical_writing.pdf}

%\end{thebibliography}

\end{document}

