\documentstyle{article}
\begin{document}

\title{Limits\\
or Seeing the Horizon\\}
\author{Kevin Iga}
\date{June 21, 1990}
\maketitle

\section{Convergent sequences}

Consider the infinite sequence

\[1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\]

Every element of the sequence is positive, but it is clear that the
sequence is approaching zero.  The sequence approaches, though it
never attains, the value zero.  When this happens, the sequence is
said to {\em converge} to zero.  But consider the following infinite
sequence:

\[1, 2, 3, 4, \ldots\]

It is clear that it will zoom off to infinity.  The sequence will {\em
not} converge to a number, but is said to {\em diverge}.  There are
many cases that are ambiguous now (like sequences that oscillate) but
we wait on those until we have a more formal definition of a limit.

We can write a sequence as

\[a_1, a_2, a_3, \ldots\]

then give a rule for their assignment, like

\[a_n = \frac{1}{n}\]

This allows us to be more precise, since we can specify the entire
infinite sequence.  After all, it is plausible that when I write $1,
2, 3, \ldots$ I mean $1, 2, 3, 4, 5, 923, 6, \ldots$, though it is not
likely since the ellipsis ($\ldots$) suggests that I have conveyed a
pattern in the numbers I have supplied, and that pattern does (\em
not} jump suddenly from 5 to 923 then return to 6.

We do not, however, want the structure of mathematics to depend on the
quirks in our pattern recognition algorithm, so the $a_n=f(n)$ gives a
convenient method (when computable in closed form) to write a sequence
exactly without using an infinite amount of pencil lead!  For
illustration, however, it is still useful to list the first few terms
and see if we {\em can} recognize a pattern.

We will often refer to the sequence with $a_n=f(n)$ as simply f(n).
So our first example tells us that the sequence $\frac{1}{n}$
converges to zero, while the second example says that $n$
diverges.

\subsection{Formal definition of a limit of a sequence}

Note that if a sequence converges to a point, it gets arbitrarily
close to that point.  In other words, you can't say that $\frac{1}{n}$
is of distance .3 away from zero, since there are points which are
closer to zero than that.  You can't say it is .0001 away from zero,
since there are points closer to zero than that.  You can't say that
it is .00000000001 away from zero, either, for the same reason.  Get
the picture?  Choose a distance (greater than zero).  No matter what
distance from zero you choose, the sequence gets closer to zero than
that.

Actually, the result is much stronger.  Not only does there exist a
point closer to zero than, say, .02 units, but ALL the points from the
fifty-first term on are all closer to zero than .02 units.  It's like
a black hole effect.  Not only do the points get arbitrarily close to
it; the sequence itself, after a certain point, never leaves the
region within any given distance from the point.

Let's call this distance epsilon ($\epsilon$), and the ``point of no
return'' in the sequence we'll call N.  The definition goes like this:

A sequence $a_1, a_2, \ldots, a_n, \ldots$ converges to $L$ if and only
if for every $\epsilon > 0$ there exists an integer $N$ such that for
every $n > N$, $|a_n-L|<\epsilon$.

This is a rather difficult definition.  It's actually a restatement of
the last paragraph.  It's really a test for if $a_n$ converges to L.
It doesn't say it must converge to some number, nor does it say
(though it can be proven) that it can only converge to one number.

You may want to think of this in terms of a game.  We get a sequence
and a number L we want to test.  You start by picking an $\epsilon >
0$.  My job is to find an N so that $a_{N+1}, a_{N+2}, \ldots$ all are
closer to L than your number.

So let's say our sequence is $\frac{1}{n}$, and we want to see if it
converges to 0.  You choose $\epsilon=.2$.  ``That's easy,'' I say, ``Just
look at the sequence from the sixth term on.  All of those are within
distance .2''.  Then you choose $\epsilon=.009$.  I then choose $N=101$.
You choose .0001, I choose 10000.  If I always win, then we know that
$a_n$ converges to L.  If you win {\em just once,} $a_n$ does not
converge to L.

For instance, let's say we chose the same sequence but took L to be 1.
When you choose $\epsilon=1$, I choose $N=1$ and I win.  But if you
choose $\epsilon=.5$, I'm stuck.  Sure, $a_1=1$, but the sequence goes
away from that point.  I can't find a starting place after which {\em
all} terms are closer to 1 than $\epsilon=.5$.  You win.  The sequence
does not converge to 1.

Look at the definition again.  It will take some time to sink in.  You
may want to read on then go back to the definition once you've played
with convergence a little.

\subsection{A simple example}

Suppose we have the sequence

\[0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots, \frac{n-1}{n},
\ldots\]

It seems clear that this will converge to one.  We want to prove this.
The way we do this is we design an algorithm for choosing $N$ if we know
$\epsilon$.  If $N$ is the next integer after $\frac{1}{\epsilon}$, we
should be safe.  In other words,

\begin{eqnarray*}
|\frac{n-1}{n}-1| & = & |\frac{n}{n} - \frac{1}{n} - 1|\\
 & = & |1 - \frac{1}{n} - 1| \\
 & = & |-\frac{1}{n}!\\
 & = & \frac{1}{n}\\
 & < & \frac{1}{N}\\
 & < & \epsilon\\
\end{eqnarray*}

So our criterion is satisfied.  Indeed, the sequence DOES converge to one.

Suppose now that we want to check that it doesn't converge to anything
else.  (Remember that our definition doesn't say anything about there
not being more than one L!)  Let's check $L\neq1$.  Note that we want
to prove that the sequence does {\em not} converge to L, so we switch
sides on our game.  I want to choose an $\epsilon$ so bad that all
hopes for finding a suitable N are dashed.  I especially don't want 1
to be in the allowable range, since after a certain point, all $a_n$
go near 1.  So I choose $\epsilon=\frac{|L-1|}{3}$, just to be safe.

Now look at $k=\frac{3}{\epsilon}=\frac{9}{|L-1|}$.  I have arranged
it so that $a_k$ is not in \epsilon, and neither is anything after it,
because

\begin{eqnarray*}
|\frac{n-1}{n}-L| & = & |\frac{1}{n}+L-1|\\
 & \leq & |\frac{1}{n}| + |L-1| \\
 & = & \frac{1}{n} + 3 \epsilon \\
 & > & \frac{1}{k} + 3 \epsilon \\
 & = & 6 \epsilon \\
 & > & \epsilon\\
\end{eqnarray*}

Note that the second line follows from the fact that you can't shorten
the distance from a to c by detouring through b $|a-c| \leq |a-b| +
|b-c|$.  Think about it if you're bothered by it; otherwise, just
memorize it as it is incidental to the topic on convergence.

I now have an $\epsilon$ and a $k$ so that the sequence after $k$
keeps away from $L$ by at least $\epsilon$ (and lots of room to
spare)!  But that's not quite the game.  I don't really get to choose
$N$.  You do.

But if $N<k$, then I have won, since it must be closer than $\epsilon$
for all $n>N$, and $k>N$.  Now if $N>k$, then I still win, since it
must be closer than $\epsilon$ for all $n>N>k$, which is true.

All of this was a long and tiresome argument that proves something
rather simple.  It's also a little ugly (or so I think) since it has
this $\epsilon$ you cook up and sloppily engineer it so that it has
enough clearance, and so on.  Convergence proofs tend to be this bad.  We
want to have a set of general rules that will allow us prove things
more easily.

\end{document}
