The Fourth Lesson on Analyis
Null Sequences in the Theory of Limits
last updated 1jul13, previously 5apr11, 5apr12\begin{document} \maketitle
Question 1.
Calculate $\gamma_n$ from the equations, and identify each rule used to
show that $\gamma_n \rightarrow 0 $.
For the product, you'll need the distributive law. Then you'll need that
a constant multiple of a null sequence is null, that the product of two
null sequences is null, and the sum of any finite number of null sequences
is null. These in turn follow nicely from the definition, using epsilons.
\subsection{Quotients of Sequences}
Let's see what you have to do with a quotient. The first step is always
mechanical:
\[ \frac{a_n}{b_n} = \frac{L+\alpha_n}{M+\beta_n} =
\frac{L}{M + \beta_n} + \frac{\alpha_n}{M+\beta_n}.\]
The RHS has two summands, and they are treated differently. But first
let's take care of the denominator. We need to assume that $M\neq 0$.
So first we must make $n > N_1$ so that $(\forall n>N_1)(\beta_n < |M|/2)$
where $N_1$ is guaranteed to exist by half the hypotheses (which half?).
This keeps the denominators on the RHS from being zero by keeping it
larger than $|M|/2$.
We can now dispose of the \textit{second} summand, by choosing an $N_2$ so
that
\[(\forall n > N_2)(|\alpha_n| < \frac{|M|}{2}\frac{\epsilon}{2}).\]
(What good is that?) This leaves the first summand and we're still not done.
We want to show that $\frac{L}{M+\beta_n}= \frac{L}{M} + \gamma_n $ for
some null sequence $\gamma_n$. So solve for
$$\gamma_n = -\frac{L}{M}\frac{\beta_n}{M+\beta_n}.$$
Check my arithmetic. And now, pick a (possibly bigger) $N_3$ for $\beta_n$
so that that $|\gamma_n| < \frac{\epsilon}{2} $. (More arithmetic.)
\textbf{ Moral of the Story: } The calculus of null functions is a nearly
mechanical way of proving tricky limit statements.
\textbf{Proof of [14.7]:} This one is actually a statement about null functions
and generalizes the idea of multiplying $0k=0$ to a null function times a
bounded function.
Question 2.
Show that if $\alpha_n\rightarrow 0$ and $|b_n|< B$ then $\alpha_n b_n$ is null.
Hint: Choose the $N_\alpha$ for $\epsilon/B$ and complete the arithmetic.
\textbf{Theorem [14.10]:}
If $0\le x \lt 1$ and $\frac{b_{n+1}}{b_n} = x + \beta_n $ then
$b_n$ is a null sequence.
\textbf{Proof:}
This is an interesting application of two concepts. First, suppose
that $b_n = L + \beta_n$, i.e. that the given sequence converges, and
$L \ne 0$. Let's,
for simplicity, assume everything is positive, which eliminates a lot
of absolute value signs. Then substituting, and applying the rules of
null functions, yields
\[ \frac{b_{n+1}}{b_n}=\frac{L+\beta_{n+1}}{L+\beta_n}}=
\frac{L}{L+\beta_n} +\frac{\beta_{n+1}}{L+\beta_n} \rightarrow 1 .\]
Note the denonminators are all bounded below by $L/2$ once the
$|\beta_n < L/2$. That $L/2$ was our epsilon for the "eventually".
Since this contradicts the hypothesis, we're done. $L=0$ and $b_n$
is null.
But we are not told to assume that $b_n$ converges. That comes first.
For $x<1$ we can insure that $\frac{b_{n+1}}{b_n} < 1 $ eventually.
But multiplying through establishes the $b_n$ to be a monotonically
decreasing sequence bounded below by $0$.
Question 3.
Explain why the limit $x$ of $b_{n+1}/b_n$ needs to be in $[0,1)$ for
$b_n$ to converge using the above argument.
\section{Preview of Null Functions}
[This section is optional and will not be tested on the final in MA348SP11.]
This brings us up to the high point of this chapter, the
\textit{ Bolzano-Weierstrass Theorem }, which we'll take up in the next
edition of these notes.
But anticipating Ch15, we can define \textit{null functions} to be
functions with this property:
\[ (\forall \epsilon> 0)(\exists \delta > 0)(\forall
- \delta < h < 0 \mbox{ and } 0 < h < \delta )( |\alpha(h)| < \epsilon)\]
The region $\{ h | -\delta < h < 0 \mbox{ and } 0 < h < \delta \}$ is
called a \textit{ punctured neighborhood} of $0$.
This allows us to work with the practical expression
\[ f(a+h) = f(a) + \alpha(h) \mbox{ for } \lim_{x\rightarrow a} f(x) = L .\]
Exercise 1. See how many of the above theorems above about convergent
sequences carries over, statement \textit{ and} proof, to functions.
A function is said to be \textit{ continuous} at $x$ if in a punctured
neighborhood of $0$ for $h$ we have that $f(x+h)=f(x) + \theta(h)$,
for some null function $\theta(h)$.
A function is said to be \textit{differentiable} at $x$, with
derivative $f'(x)=m$, if in a
punctured neighborhood of $0$ for $h$ we have that
\[ f(x+h) = f(x) + (m + \theta(h))h \] for some null function.
Exercise 2. See how many properties of derivatives you can now
{\it calculate} directly from this definition.
If you'll do the sum and product rules of differentiation, then I'll do the
chain rule, which is the most remarkable improvement over the usual
exposition in standard calculus books.
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