\documentclass{ymexam}

\usepackage{times}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}

\newcommand{\eps}{\varepsilon}

\begin{document}

%\resit
\codenumber{0500010}
\year{2006}
\title{Analysis I}
\duration{$1\half$ hours}

\begin{rubric}
  Answer \underline{all} questions in Part A (questions 1--5) and
  \underline{two} questions from Part B (questions 6--8). \\
  Part~A carries 40~marks; questions in part~B carry 25 marks each. \\
  Please write your answers in ink; pencil is acceptable for graphs and
  diagrams. Do not use red ink.\\
  Standard calculators will be provided. \\
  The marking scheme shown on each question is indicative only.
\end{rubric}

\maketitle

\part{Answer all questions in this part of the paper}

\begin{question}
  Suppose $S$ is a non-empty subset of $\R$. Define the terms
  \emph{supremum} and \emph{infimum}, as applied to $S$. Under what
  circumstances do the supremum and infimum exist?
    
  For each of the following sets, write down the supremum and infimum, 
  if they exist. \emph{You are not required to justify your answers}.
  $$S_1=\{x\in\R:(x-1)^2<3\}; \qquad  S_2=\{x\in\R:x^3\leq 2\}.$$
  \marks{8}
\end{question}

\begin{question}
  State the definition of \emph{convergence} of a sequence of real 
  numbers $(a_n)_{n\in\N}$ to a limit $a\in\R$, and use this
  definition to show that
  $$\frac{2n}{n+3}\to 2\text{ as }n\to\infty.\marks{10}$$
\end{question}

\begin{question}
  Determine the limits as $n\to\infty$ of the following sequences.
  \emph{Make free use of standard combination rules and standard limits,
    briefly indicating which results you are using; ``$\eps$''
    arguments are \textbf{not} required.}
    $$a_n=\frac{2-n+n^2}{n^2+3}; \qquad
      b_n=\frac{3^n-n}{4^n+\sqrt{2}}; \qquad
      c_n=\frac{n!+n}{(n+1)^n}.$$
    \marks{6}
\end{question}

\begin{question}
  Determine which of the following series converge:
  \begin{enumerate}
  \item $\displaystyle\sum_{n=1}^\infty\frac{n!}{x^n}$ ($x\neq 0$ fixed);
  \item $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{\log(n+1)}$.\marks{6}
  \end{enumerate}
\end{question}

\begin{question}
  What is meant by the \emph{radius of convergence} of a power series
  $$\sum_{n=0}^\infty a_n(x-x_0)^n$$
  (including the cases $0$ and $\infty$)? Find the radius of
  convergence of the following series:
  \begin{enumerate}
  \item $\displaystyle\sum_{n=1}^\infty\frac{(x+2)^n}{\sqrt{n}}$;
  \item $\displaystyle\sum_{n=1}^\infty\frac{x^n}{n^{n/2}}$.\marks{10}
  \end{enumerate}
\end{question}

\clearpage
\part{Answer two questions from this part of the paper}

\begin{question}
  Suppose $S$ is a bounded, non-empty subset of $\R$. Define the terms
  \emph{upper bound} and \emph{lower bound}, as applied to $S$.\marks{4}
  
  Let
  $$-S=\{-x:x\in S\}.$$
  Show that $b$ is an upper bound for $S$ if and only if $-b$ is a lower
  bound for $-S$, and that $\inf(-S)=-\sup(S)$. Deduce that
  $\sup(-S)=-\inf(S)$.\marks{14}
  
  State without proof the \emph{Archimedean property} (or \emph{axiom of
    Archimedes}) and use it to show that
  $$\inf\{1/n:n\in\N\}=0.$$
  What is the supremum of this set? Justify your answer.\marks{7}
\end{question}

\begin{question}
  \emph{Throughout this question, make free use of the standard combination 
    rules and the basic fact that $1/n\to 0$ as $n\to\infty$.}

  State without proof the \emph{Sandwich Theorem}.\marks{3}

  Show by induction that $2^n>n$ for all $n\in\N$, and deduce that $1/2^n\to 0$
  as $n\to\infty$.\marks{6}
  
  By expanding the numerator and denominator into $n$ factors, show that
  if $n\geq 3$ then
  $$\frac{2^n}{n!}\leq\frac{4}{n}.$$
  Deduce that $2^n/n!\to 0$ as $n\to\infty$.\marks{8}
  
  Apply a similar technique to show that $(n!)^2/(2n)!\to 0$ as $n\to\infty$.
  \marks{8}
\end{question}

\begin{question}
  \emph{Throughout this question, make free use of any standard facts about 
    convergence of geometric series.}

  State without proof the \emph{comparison test}.\marks{4}
  
  Suppose $(a_n)_{n\in\N}$ is a sequence such that for every $x\in\R$
  with $|x|<1$, the sequence $(a_nx^n)_{n\in\N}$ is bounded; say
  $|a_nx^n|\leq C_x$ for all $n\in\N$.

  Suppose $0\leq x<1$. By considering the factorisation and estimate 
  $$|a_nx^n|=|a_nx^{n/2}|x^{n/2}\leq C_{\sqrt{x}}x^{n/2},$$
  or otherwise, show that the series
  $$\sum_{n=1}^\infty a_nx^n$$
  is absolutely convergent.\marks{10}
  
  Show that the same is true if $-1<x<0$. Express these results in terms
  of the radius of convergence of a power series.\marks{11}
\end{question}

\begin{solution}
  The supremum of $S$ is the least upper bound of $S$, if 
  $S$ is bounded above. The infimum is the greatest lower bound 
  of $S$, if $S$ is bounded below.\marks{4}
    
  $$\sup(S_1)=1+\sqrt{3}; \qquad \inf(S_1)=1-\sqrt{3}$$
  $$\sup(S_2)=2^{1/3}; \qquad \inf(S_2)\text{ does not exist}\marks{4}$$
\end{solution}

\begin{solution}
  $a_n$ converges to  $a$ as $n\to\infty$ if for any $\eps>0$
  there exists $N_\eps\in\N$ such that if $n>N_\eps$ then $|a_n-a|<\eps$.
  \marks{4}
  If $a_n=2n/(n+3)$ and $a=1$ then we calculate:
  \begin{align*}
    |a_n-a|<\eps &\iff |2n/(n+3)-2|<\eps \\
                 &\iff |-6/(n+3)|<\eps \\
                 &\iff 6/(n+3)<\eps \\
                 &\iff n>6/\eps-3
  \end{align*}    
  Given $\eps>0$, let $N_\eps$ be some natural number greater than
  $6/\eps-3$. If $n>N_\eps$ then $n>6/\eps-3$ so by the above
  calculation (all steps being ``if and only if'') it follows that
  $|a_n-a|<\eps$, showing that $a_n\to a$ as $n\to\infty$.\marks{6}
\end{solution}

\begin{solution}
  $$a_n=\frac{2-n+n^2}{n^2+3}=\frac{2/n^2-1/n+1}{1+3/n^2}\to\frac{0-0+1}{1+0}=1$$
  using standard combination rules and $1/n,1/n^2\to 0$.
  $$b_n=\frac{3^n-n}{4^n+\sqrt{2}}=\frac{(3/4)^n-n/4^n}{1+\sqrt{2}/4^n}\to
        \frac{0-0}{1+0}=0$$
  using combination rules, $(3/4)^n,(1/4)^n\to 0$ since $|3/4|<1,|1/4|<1$
  and ``exponential beats power''.
  $$c_n=\frac{n!+n}{(n+1)^n}=\frac{n!}{(n+1)^n}+\frac{n}{(n+1)^n}\to 0+0=0$$
  using combination rules and ``$n^n$ beats factorial''.\marks{6}
\end{solution}

\begin{solution}
  \begin{enumerate}
  \item Ratio test:
    $$\left|\frac{(n+1)!}{x^{n+1}}\frac{x^n}{n!}\right|=\frac{n+1}{|x|}$$
    This tends to $+\infty$ for any $x\neq 0$. so the series diverges.
    \marks{3}
  \item $1/\log(n+1)$ decreases and tends to zero, so by Leibniz's
    alternating series test, the series converges.\marks{3}
  \end{enumerate}
\end{solution}

\begin{solution}
  The radius of convergence of the given power series is $0$ if it
  converges only when $x=x_0$, or $\infty$ if it converges for all $x\in\R$.
  Otherwise, it is the number $R>0$ with the property that the series
  converges if $|x-x_0|<R$ and diverges if $|x-x_0|>R$.\marks{4}

  In both cases we use the ratio test.

  \begin{enumerate}
  \item $$\left|\frac{(x+2)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+2)^n}\right|
    =|x+2|\sqrt{n/(n+1)}\to|x+2|$$
    This show that we have convergence if $|x+2|<1$ and divergence if
    $|x+2|>1$; that is, $R=1$.\marks{3}
   \item $$\left|\frac{(x+1)^{n+1}}{(n+1)^{(n+1)/2}}\frac{n^{n/2}}{(x+1)^n}\right|=
    \frac{|x+1|}{\sqrt{n+1}}\left(\frac{n}{n+1}\right)^{n/2}\to 0$$
    using the fact that $n/(n+1)<1$. Since the limits is $<1$ for all
    $x$, we have convergence for all $x$; that is, $R=\infty$. \marks{3}
  \end{enumerate}
\end{solution}

\begin{remarks}
  Everything in Section A is supposed to be straightforward and,
  if not identical to examples they've already seen, certainly
  similar. Full credit, of course, for equally correct solutions, e.g.
  root test in very last part.
\end{remarks}
  

\begin{solution}
  An upper bound of $S$ is a number $b\in\R$ such that for all $x\in S$,
  $x\leq b$. A lower bound is a number $a\in\R$ such that for all 
  $x\in S$, $a\leq x$.\marks{4}

  $b$ is an upper bound for $S$ $\iff$ for all $x\in S$, $x\leq b$ $\iff$
  for all $x\in S$, $-x\geq -b$ $\iff$ for all $y\in-S$, $-b\leq y$ $\iff$
  $-b$ is a lower bound for $-S$.\marks{4}

  Since $\sup(S)$ is an upper bound for $S$, $-\sup(S)$ is a lower bound
  for $-S$. If $a>-\sup(S)$ then $-a<\sup(S)$, so $-a$ is not an upper
  bound for $S$; it follows that $a$ is not a lower bound for $-S$. This
  shows that $-\sup(S)$ is the greatest lower bound of $-S$; that is,
  $\inf(-S)=-\sup(S)$.\marks{6}
  
  To see that $\sup(-S)=-\inf(S)$, it is possible to use a similar
  argument; but it is easier to observe that $-(-S)=S$ so, appying the
  above result to $(-S)$, we have $\inf(S)=\inf(-(-S))=-\sup(-S)$, or
  equivalently $\sup(-S)=-\inf(S)$.\marks{4}
  
  Archimedes axiom states that the natural numbers are not bounded above;
  that is, for any $x\in\R$ there exists $n\in\N$ such that $n>x$.
  Clearly $0$ is a lower bound of the given set. If $a$ is another lower
  bound, then $a\leq 1/n$ for all $n\in\N$. In case $a>0$ this implies
  $n<1/a$ for all $n\in\N$, in contradiction to Archimedes. We therefore
  have $a\leq 0$, so $0$ is the greatest lower bound.\marks{5}

  Since $1\leq n$ for all $n\in\N$, $1/n\leq 1$ for all $n\in\N$. This
  shows that $1$ is an upper bound of the given set. But $1=1/1$ is also
  an element of the set, so it is the maximum and, in particular, the supremum.
  \marks{2}
  
  \begin{remarks}
    Candidates typically find formal calculations with bounds difficult,
    so I've tried to make this question reasonably straightforward. The
    calculation $\inf(S)=-\sup(-S)$ is in the notes, in the context if
    proving that infima exist. The infimum of $\{1/n:n\in\N\}$ is also
    in the notes, as part of a list of equivalent formulation of the
    Archimedean proper, but the proof isn't as direct as this.
  \end{remarks}

\end{solution}

\begin{solution}
  The Sandwich Theorem states that if $(a_n)_{n\in\N}$, $(b_n)_{n\in\N}$ and 
  $(c_n)_{n\in\N}$ are real sequences such that $a_n\leq b_n\leq c_n$ for 
  all $n$ and $a_n$ and $c_n$ both converge to the same limit $a$ as
  $n\to\infty$, then $b_n\to a$ as $n\to\infty$. \marks{3}

  Certainly $2^1>1$. If, for some $n$, $2^n>n$, then 
  $$2^{n+1}=2.2^n>2n=n+n\geq n+1.\marks{3}$$
  It follows that $2^n>$ for all $n$. We now have
  $$0<\frac{1}{2^n}<\frac{1}{n}$$
  so $1/2^n\to 0$ as $n\to\infty$ by the sandwich theorem.\marks{3}
  
  We have
  $$\frac{2^n}{n!}=\frac{2}{1}\frac{2}{2}\frac{2}{3}\frac{2}{4}\dots\frac{2}{n}$$
  The product of the first two terms is $2$. The next $n-3$ terms are all 
  less than $1$. The last term is $2/n$. This gives
  $$0<\frac{2^n}{n!}\leq 2\frac{2}{n}=\frac{4}{n}$$
  for $n\geq 3$.\marks{5}
  The RHS tends to $0$, so $2^n/n!\to 0$ as $n\to\infty$ by the sandwich 
  theorem (the cases $n=1,2$ not making any difference to the limit).\marks{3}
  
  Similarly, the first $n$ terms in the expansion $(2n)!=1.2\dots (2n)$
  cancel with $n!$ to give
  $$\frac{(n!)^2}{(2n)!}=\frac{1}{n+1}\frac{2}{n+2}\dots\frac{n}{2n}$$
  and each term on the RHS is $\leq 1/2$. This gives
  $$0<\frac{(n!)^2}{(2n)!}\leq\frac{1}{2^n}\marks{5}$$
  The RHS tends to $0$ by an earlier result so $(n!)^2/(2n)!\to 0$ as
  $n\to\infty$ by the sandwich theorem.\marks{3}
  
  \begin{remarks}
    Inequalities like this, and the consequences of the sandwich theorem, are
    familiar, though mostly in more general contexts such as $x^n/n!\to 0$ for 
    any $x$. The last inequality was on a problem sheet.
  \end{remarks}
\end{solution}

\begin{solution}
  Suppose $(p_n)_{n\in\N}$ and $(q_n)_{n\in\N}$ are series such that
  $|p_n|\leq|q_n|$ for all $n$ and $\sum_{n=1}^\infty q_n$ is absolutely
  convergent. Then $\sum_{n=1}^\infty p_n$ is absolutely convergent.\marks{4} 

  The suggested factorisation 
  $$a_nx^n=a_nx^{n/2}x^{n/2}=a_n(\sqrt{x})^n(\sqrt{x})^n$$
  yields
  $$|a_nx^n|\leq C_{\sqrt{x}}(\sqrt{x})^n$$
  Since $0<x<1$, we also have $0<\sqrt{x}<1$ so the geometric sum 
  $\sum_{n=1}^\infty(\sqrt{x})^n$ converges. By the comparison test,
  $\sum_{n=1}^\infty a_nx^n$ is absolutely convergent.\marks{10}
  
  In the case $-1<x<0$, we can instead use the factorisation
  $$a_nx^n=a_n(-1)^n|x|^{n/2}|x|^{n/2}=a_n(-1)^n(\sqrt{|x|})^n(\sqrt{|x|})^n$$
  which leads to
  $$|a_nx^n|\leq C_{\sqrt{|x|}}(\sqrt{|x|})^n$$
  The comparison test now works in the same way to show that 
  $\sum_{n=1}^\infty a_nx^n$ is absolutely convergent, since $0<\sqrt{|x|}<1$.
  \marks{8}
  
  This hows that the power series $\sum_{n=1}^\infty a_nx^n$ has radius
  of convergence $R\geq 1$.\marks{3}
  
  \begin{remarks}
    Apart from the comparison test and the convergence of the geometric
    series, this is unseen. It's quick and easy for the good students.
  \end{remarks}
\end{solution}
    
\end{document}

