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\title{MTH435 - Homework 6}



\begin{document}
 \maketitle
 
 Solutions to the questions in Section B should be submitted by the start of class on 10/31/18.
 
 
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 \section{Warm-up Questions}
 
 \begin{que}
  Decide if the following sequences are properly divergent.
  \begin{enumerate}
   \item $(\sqrt{n})$.
   \item $(n \cos n)$
   \item $\left( \frac{2n}{\sqrt{n+7}}\right)$
   \item $\left( \frac{n+1}{\sqrt{n^2+5}}\right)$
  \end{enumerate}

 \end{que}

 \begin{que}
  Compute the following.
  \begin{enumerate}
   \item $\sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)}$.
   \item $\sum_{n=1}^{\infty} \frac{1}{n(n+2)}$.
   \item $\sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^{3n}$.
  \end{enumerate}

 \end{que}

 
\begin{que}
 Give examples of the following.
 \begin{enumerate}
  \item A properly divergent sequence that is not monotonic.
  \item An unbounded sequence that is not properly divergent.
  \item An unbounded sequence with a convergent subsequence.
 \end{enumerate}

\end{que}

 \begin{que}
  Let $(x_n)$ be a sequence of non-zero real numbers. Prove or disprove the following.
  \begin{enumerate}
   \item If $x_n \to 0$ then $\frac{1}{x_n}$ is properly divergent.
   \item If $x_n \to +\infty$ then $\frac{1}{x_n} \to 0$.
  \end{enumerate}

 \end{que}

\begin{que}
 Prove or disprove the following.
 \begin{enumerate}
  \item If $x_n \to +\infty$ and $(y_n)$ is convergent then $(x_n+y_n) \to + \infty$.
  \item If $x_n \to +\infty$ and $(y_n)$ is convergent then $(x_ny_n) \to +\infty$.
  \item If $x_n \to +\infty$ and $(x_ny_n)$ converges then $(y_n)$ converges.
  \item If $x_n \to +\infty$ and $y_n \to +\infty$ then $(x_n - y_n)$ converges.
  \item If $\sum a_n$ converges then $\sum a_n^2$ converges. What if $a_n > 0$ for all $n$?
  \item If $\sum a_n$ converges and $a_n > 0$ for all $n$ then $\sum \sqrt{a_n}$ converges.
  \item If $(a_n)$ is a decreasing sequence of positive numbers and if $\sum a_n$ converges then $\displaystyle \lim_{n \to \infty} na_n = 0$.
  \item If $(a_n)$ is such that $(n^2 a_n)$ converges then $\sum a_n$ is absolutely convergent.
 \end{enumerate}

\end{que}
 
 
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 \section{Submitted Questions}
 
 \begin{que}
Let $(x_n)$ and $(y_n)$ be sequences of positive numbers and suppose $\left( \frac{x_n}{y_n}\right) \to +\infty$.
  \begin{enumerate}
  \item Show that if $y_n \to +\infty$ then $x_n \to +\infty$.
  \item Show that if $(x_n)$ is a bounded sequence then $(y_n)$ converges.
\end{enumerate}
 \end{que}

 \begin{que} \mbox{}
  \begin{enumerate}
   \item Show that if $\sum a_n$ is absolutely convergent and $(b_n)$ is a bounded sequence then $\sum a_n b_n$ is absolutely convergent.
   \item Give an example of a conditionally convergent series $\sum a_n$ and a bounded sequence $(b_n)$ such that $\sum a_n b_n$ is divergent.
  \end{enumerate}

 \end{que}

 
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 \section{Challenge Questions}
 
  \begin{que} \emph{Another definition of $e$}. We give an alternative definition of $e$ and prove it is irrational. First, consider the sequence $(e_n)$ defined by
 \[
  e_n = \sum_{k=0}^n \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dotsb + \frac{1}{n!}.
 \]
\begin{itemize}
 \item[(i)] Define $s_n = 1 + 1 + \frac{1}{2} + \frac{1}{4} + \dotsb + \frac{1}{2^{n-1}}$ for $n \geq 1$. Prove that $s_n \geq e_n$ for all $n$ and deduce that $(e_n)$ converges. We denote this limit by $e$.
 \item[(ii)] Assume $e = p/q$ in lowest terms and show that $e - e_q = \frac{k}{q!}$ for some $k \in \N$. 
 \item[(iii)] Prove that $e - e_q < \frac{1}{q!} \sum_{k=1}^\infty 2^{-k}$ and obtain a contradiction to part (ii). Conclude that $e$ is irrational.
\end{itemize}
 \end{que}

 \begin{que} We now prove that the new definition of $e$ given in C.1. is equivalent to the definition given in class. Denote
  \[
   e = \sum_{n=0}^{\infty} \frac{1}{n!}.
  \]
\begin{itemize}
 \item[(i)] Use the binomial theorem to prove that
 \[
  \left( 1 + \frac{1}{n} \right)^n = \sum_{k=0}^n \frac{1}{k!} \left(1 - \frac{1}{n} \right)\left( 1 - \frac{2}{n} \right) \dotsm \left( 1 - \frac{(k-1)}{n} \right) \leq \sum_{k=0}^n \frac{1}{k!}
 \]
and so deduce $\displaystyle \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n \leq e$.
\item[(ii)] Use the binomial theorem to show that for all $n,m \in \N$ we have
\[
 \left( 1 + \frac{1}{n} \right)^{m+n} = \sum_{k=0}^{m+n} \frac{1}{k!} \frac{(n+m)!}{(n+m-k)!} \frac{1}{n^k} \geq \sum_{k=0}^{m} \frac{1}{k!} \frac{(n+m)!}{(n+m-k)!} \frac{1}{n^k}.
\]
Deduce that $\left( 1 + \frac{1}{n} \right)^{m+n} \geq \sum_{k=0}^m \frac{1}{k!}$.
\item[(iv)] Using the result of part (iii), show that $\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n \geq \sum_{k=0}^m \frac{1}{k!}$. Then, taking the limit as $m \to \infty$, conclude that $\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n \geq e$.
\item[(v)] Use the results of parts (i) and (iv) to prove that $\sum_{n=0}^{\infty} \frac{1}{n!} = \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)$.

\end{itemize}

\end{que}

 
 
 
 
 
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