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\title{Workshop Notes}
\author{Robert Dougherty-Bliss}
\date{10 February 2021}
\begin{document}
\maketitle
% Adding the * gets rid of the section numbering.
\section*{Logistics}%
\label{sec:logistics}
Nothing to say here.
\section*{New stuff}%
\label{sec:new_stuff}
Since I last saw you, you learned more about sequences and got some new
homework. We're going to talk about some sequence topics and a problem from the
new homework, as well as any other questions you might have.
\subsection*{Ces\`aro means}%
\label{sub:cesaro_means}
If we have a sequence $x(n)$, then we can take \emph{averages} of it by
defining
\begin{equation*}
y(n) = \frac{1}{n} \sum_{k = 1}^n x(k).
\end{equation*}
\textbf{Plausible statement:} If $x(n)$ converges, then the averages also
converge to the same thing.
This is true. It's called Ces\`aro summation. The proof illustrates one of the
main techniques of analysis: Splitting a problem into two managable parts.
\begin{theorem}
If $x(n) \to x$, then
\begin{equation*}
y(n) = \frac{1}{n} \sum_{k = 1}^n x(k) \to x.
\end{equation*}
\end{theorem}
\begin{proof}
We want to show that, given any $\epsilon > 0$, there exists some integer
$N$ such that
\begin{equation*}
|y(n) - x| < \epsilon
\end{equation*}
for $n > N$.
The trick about averaging is that we can break up a single $x$ into many
terms:
\begin{equation*}
y(n) - x = \frac{1}{n} \sum_{k = 1}^n x(k) - x = \frac{1}{n} \sum_{k = 1}^n (x(k) - x).
\end{equation*}
Therefore, by the triangle inequality,
\begin{equation*}
|y(n) - x| \leq \frac{1}{n} \sum_{k = 1}^n |x(k) - x|.
\end{equation*}
For ``large'' $k$, we have the ``sharp'' bound $|x(k) - k| < \epsilon$. For
``small'' $k$, we only have the ``coarse'' fact that $|x(k) - k|$ is
bounded, since $x(k) - k \to 0$.
To make this idea rigorous, pick $M$ such that $|x(k) - x| \leq M$ for all
$k$ (by boundedness), and $N$ such that $|x(k) - x| < \epsilon$ for $k > N$
(by convergence). Then, for $n > N$,
\begin{align*}
|y(n) - x|
&\leq \frac{1}{n} \sum_{k = 1}^N |x(k) - x| + \frac{1}{n} \sum_{N < k \leq n} |x(k) - x| \\
&\leq \frac{NM}{n} + \frac{\epsilon (n - N)}{n} \\
&\leq \frac{NM}{n} + \epsilon.
\end{align*}
To get the right bound here, we need to turn that $NM / n$ into something
involving $\epsilon$. But $N$ and $M$ are both fixed relative to
$\epsilon$, so we can just require $n$ to be really big in comparison. That
is, let $N'$ be such that $N' > N$ and $NM / n < \epsilon$ for $n > N'$.
Then, for $n > N'$, the bound becomes
\begin{equation*}
|y(n) - x| \leq \epsilon + \epsilon = 2\epsilon.
\end{equation*}
Since $\epsilon$ was arbitrary, this is equivalent and shows that $y(n) \to
x$.
\end{proof}
\textbf{Interesting question:} If the averages converge, does $x(n)$ converge?
\subsection*{Limits superior and inferior}%
\label{sub:limits_superior_and_inferior}
Not every sequence has a limit, but sometimes we want to talk about ``limits''
even when they don't exist. For example, $a(n) = (-1)^n$ does not ``converge,''
but it kind of has two ``pseudo-limits,'' $1$ and $-1$. These ``peusdo-limits''
are made precise by the \emph{limits superior and inferior}.
Given any sequence $a(n)$, there exists a number $\beta$ (possibly $\infty$)
such that:
\begin{enumerate}
\item For every $\epsilon > 0$, we have $a(n) < \beta + \epsilon$ eventually.
\item For every $\epsilon > 0$, there are infinitely many $n$ such that
$a(n)$ is in $(\beta - \epsilon, \beta]$.
\end{enumerate}
The number $\beta$ is called the \emph{limit superior} of $a(n)$, and we denote
it by $\beta = \limsup_n a(n)$.
The above properties actually uniquely define the limit superior, but they
don't prove that it exists. Let's go through how Abbott does it. For now,
suppose that $a(n)$ is a bounded sequence.
\begin{enumerate}[label=(\alph*)]
\item Prove that $y(n) = \sup \{a(k) \mid k \geq n\}$ converges. Let
$\limsup_n a(n) = \lim_n y(n)$.
% The sets are decreasing.
\item Give the analogous definition for $\liminf_n a(n)$.
% Switch to infs. Now the sets are increasing.
\item Show that $\liminf_n a(n) \leq \limsup_n a(n)$. [Hint: If $x(n)$ and
$y(n)$ converge and $x(n) \leq y(n)$, then $\lim_n x(n) \leq \lim_n
y(n)$. (You should prove this!)]
% The hint gives it away.
\item Show that $\lim_n a(n) = x$ iff $\liminf_n a(n) = x = \limsup_n
a(n)$. (You can't use the above properties I mentioned unless you prove
them!)
% Say that a(n) -> x. For every eps > 0, we have a(n) in (x - eps, x + eps)
% for large n. Thus sup {a(k) | k >= n} <= x + eps for large n, which gives
% limsup a(n) <= x + eps for every eps > 0. Thus limsup a(n) <= x.
% Same argument with flipped signs works for liminf.
% Conversely, say that limsup = liminf = x. For every eps > 0, we have
% sup {a(k) | k >= n} < x + eps eventually, so a(k) <= x + eps eventually.
% By an analogous argument, a(k) >= x - eps eventually, so a(k) is in
% [x - epsilon, x + epsilon] eventually.
\end{enumerate}
\begin{Exercise}
If $a(n) \leq b(n)$, we can't always say that $\lim_n a(n) \leq \lim_n
b(n)$, because the limits might not even exist. Show that
\begin{equation*}
\limsup_n a(n) \leq \limsup_n b(n)
\end{equation*}
and
\begin{equation*}
\liminf_n a(n) \leq \liminf_n b(n)
\end{equation*}
whenever $a(n) \leq b(n)$ for all $n$.
\end{Exercise}
\end{document}