it does not ignore \cursor and \lyxvcid
see attached
regards
john
--
"Yeah, I woke up in the day accidentally once, the moon was on fire for some
reason and I couldn't see very well and all the bandwidth disappeared, it was
very scary :("
- Orion
#This file was created by <dlj0> Mon Jul 28 23:27:10 1997
#LyX 0.11 (C) 1995-1997 Matthias Ettrich and the LyX Team
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\language american
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\quotes_language english
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\paperpagestyle headings
\layout LaTeX Title
Math 22, Second Summer Term 1997
\layout Date
\latex latex
\backslash
today
\layout Address
Lehigh University
\layout Email
[EMAIL PROTECTED]
\layout Standard
\latex latex
\backslash
maketitle
\layout Section*
Chapter 10.
Sequences and series
\layout Standard
This idea will form the basis of the rest of this course.
\layout Standard
Ominous-sounding, isn't it?
\layout Standard
I always begin a discussion of series and sequences (these two topics are
really one idea, viewed from two different sides) with Greek mythology.
This keeps alive the belief that we are fostering liberal education.
\layout Standard
Today's myth is that of Achilles and the tortoise.
The story is, for reasons that I'll never figure out, they are going to
have a race.
(This is also told with a hare in the role of Achilles, but then it doesn't
seem so Greek.) The paradox involved in this story is that, assuming Achilles
gives the tortoise a head start, which is only fair, he can never win,
since in order to cover the distance from Achilles to the tortoise (the
head start), Achilles must first go half that distance, then half the remaining
distance, and then half of that,
\shape italic
et cetera
\shape default
, so that he has to do
\shape italic
infinitely
\shape default
many things in order to catch up to the tortoise.
\layout Standard
As with all the Greek paradoxes, this has an explanation.
Let's assume that the distance from Achilles to the tortoise is 1 mile.
Then, in order to get to the tortoise, Achilles must first go a distance
1/2 mile, which would take him, say, 2 minutes.
He then has to go the next 1/4 mile, which would take 1 minute, then the
next 1/8 mile, taking 1/2 minute, then the next 1/16 mile, taking 1/4 minute,
\shape italic
et cetera
\shape default
.
\layout Standard
So, adding this all up, he has to run:
\begin_inset Formula
\[
1/2+1/4+1/8+1/16+\dots \: miles,\]
\end_inset
which takes:
\begin_inset Formula
\[
2+1+1/2+1/4+\dots \; minutes.\]
\end_inset
These totals may seem infinite, since they do go on forever, like some demented
math professor, but they actually do approach a finite value as a limit.
\layout Standard
When you think about it, Achilles only has to run the one mile (the tortoise
hasn't even taken a step yet), which is
\begin_inset Formula \( 1/2+1/4+1/8+\dots =1 \)
\end_inset
, and it takes
\begin_inset Formula \( 2+1+(1/2+1/4+\dots )=4 \)
\end_inset
minutes (he
\shape italic
is
\shape default
fast).
There is really no paradox, since the infinitely many things he has to
do can be done infinitely fast at the end.
\layout Standard
There are many oddities involved in this infinite summation process, which
we call an
\shape italic
infinite series
\shape default
.
An infinite series is an infinite sum.
There are a few examples where we really can see the infinite sum, like
the one above.
\layout Enumerate
\begin_inset Formula \( 1+1/2+1/4+1/8+\dots =2 \)
\end_inset
\layout Enumerate
Since
\begin_inset Formula \( \frac{1}{1-x}=1+x+x^{2}+x^{3}+\dots +x^{n}+\frac{x^{n+1}}{1-x}
\)
\end_inset
, which can be seen by long division, then
\begin_inset Formula \( 1+x+x^{2}+x^{3}+\dots =\frac{x^{n+1}}{1-x} \)
\end_inset
for any number
\begin_inset Formula \( x \)
\end_inset
so that
\begin_inset Formula \( \lim _{n\rightarrow \infty }\frac{x^{n+1}}{1-x}=0 \)
\end_inset
, which happens when
\begin_inset Formula \( |x|<1 \)
\end_inset
.
This is called the
\shape italic
geometric series
\shape default
.
\layout Enumerate
As an example of a geometric series:
\begin_inset Formula \( 1+1/3+1/9+1/27+\dots =1/(1-(1/3))=3/2 \)
\end_inset
.
\layout Enumerate
\begin_inset Formula \( \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots
=1 \)
\end_inset
.
\layout Standard
The last series, called the
\shape italic
telescoping series
\shape default
, is the easiest of all to see, once you know the trick:
\begin_inset Formula \( \frac{1}{n\cdot (n+1)}=\frac{1}{n}-\frac{1}{n+1} \)
\end_inset
.
So, you add them together:
\begin_inset Formula
\begin{eqnarray*}
\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots & = &
(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\dots
\\
& = &
\frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})+(-\frac{1}{3}+\frac{1}{3})+(-\frac{1}{4}+\frac{1}{4})+(-\frac{1}{5}+\dots
)+\dots \\
& = & 1.
\end{eqnarray*}
\end_inset
\layout Section*
\latex latex
\backslash
S
\latex default
10.1, Sequences
\layout Standard
We need to get some terminology straight here.
A
\shape italic
series
\shape default
, or
\shape italic
infinite series
\shape default
, is the infinite sum.
The things you add up are the
\shape italic
terms
\shape default
of the series.
They form an
\shape italic
infinite sequence
\shape default
.
A sequence is just the listing, one number after another.
A series is a summing of those numbers.
\layout Standard
Mathematicians think of sequences as being easier to understand.
Just one number after another.
You can usually tell if the sequence is approaching something, like
\begin_inset Formula \( 1/n \)
\end_inset
approaches 0, as
\begin_inset Formula \( n \)
\end_inset
goes to
\begin_inset Formula \( \infty \)
\end_inset
, using l'H
\latex latex
\backslash
^{o}
\latex default
pital's rule and similar ideas.
Technically, a sequence is a function of the positive integers, but we
write one as
\begin_inset Formula \( \{a_{n}\} \)
\end_inset
.
We say that a sequence
\begin_inset Formula \( \{a_{n}\} \)
\end_inset
has a
\shape italic
limit
\shape default
\begin_inset Formula \( L \)
\end_inset
as
\begin_inset Formula \( n\rightarrow \infty \)
\end_inset
, if the numbers (
\shape italic
terms
\shape default
) approach
\begin_inset Formula \( L \)
\end_inset
as
\begin_inset Formula \( n \)
\end_inset
gets large.
\shape italic
Really
\shape default
technically, we say:
\layout Definition*
A sequence
\begin_inset Formula \( \{a_{n}\} \)
\end_inset
has a limit
\begin_inset Formula \( L \)
\end_inset
if, for any
\begin_inset Formula \( \epsilon >0 \)
\end_inset
, there is a
\begin_inset Formula \( N\gg 0 \)
\end_inset
so that, if
\begin_inset Formula \( n>N \)
\end_inset
,
\begin_inset Formula \( |a_{n}-L|<\epsilon \)
\end_inset
.
\layout Standard
Now that that's out of my system, let's look at some examples.
We usually really only care about
\shape italic
limits
\shape default
of sequences, and whether they exist.
For example, find the limits of the following sequences, if they exist:
\layout Enumerate
\begin_inset Formula \( a_{n}=1/2^{n} \)
\end_inset
\layout Enumerate
\begin_inset Formula \( b_{n}=2^{n} \)
\end_inset
\layout Enumerate
\begin_inset Formula \( c_{n}=(1+\frac{1}{n})^{n} \)
\end_inset
(This
\shape italic
should
\shape default
remind you of something).
\layout Enumerate
\begin_inset Formula \( d_{n}=sin(n) \)
\end_inset
.
\layout Standard
A tougher example is given by the following list:
\layout Enumerate
\begin_inset Formula \( a_{n}=2^{n}/n! \)
\end_inset
\layout Enumerate
\begin_inset Formula \( b_{n}=k^{n}/n! \)
\end_inset
\layout Enumerate
\begin_inset Formula \( c_{n}=n!/n^{n} \)
\end_inset
.
\layout Subsection*
Some Theoretical Results
\layout Theorem
\begin_inset LatexCommand \label{r^n}
\end_inset
If
\begin_inset Formula \( -1<r<1 \)
\end_inset
,
\begin_inset Formula
\[
\lim _{n\rightarrow \infty }r^{n}=0.\]
\end_inset
Also, if
\begin_inset Formula \( r\leq -1 \)
\end_inset
or
\begin_inset Formula \( r>1 \)
\end_inset
,
\begin_inset Formula
\[
\lim _{n\rightarrow \infty }r^{n}\, \mathrm{Does}\, \mathrm{Not}\, \mathrm{Exist}.\]
\end_inset
\layout Proof
We can at least show that the limit would have to be 0, if it exists.
To show it exists, we need the next result.
But, if
\begin_inset Formula
\[
\lim _{n\rightarrow \infty }r^{n}=L,\]
\end_inset
which is
\shape italic
some
\shape default
number, maybe not 0, then
\begin_inset Formula
\[
L=\lim _{n\rightarrow \infty }r^{n}=\lim _{n\rightarrow \infty }r^{n+1}=\lim
_{n\rightarrow \infty }(r^{n})r=(\lim _{n\rightarrow \infty }r^{n})r=Lr.\]
\end_inset
Since
\begin_inset Formula \( L=Lr \)
\end_inset
, but
\begin_inset Formula \( r\neq 1 \)
\end_inset
,
\begin_inset Formula \( L \)
\end_inset
has to be 0.
\layout Standard
To show the limit exists, for
\begin_inset Formula \( -1<r<1 \)
\end_inset
, and doesn't converge for
\begin_inset Formula \( r\leq -1 \)
\end_inset
or
\begin_inset Formula \( r>1 \)
\end_inset
, we need to talk about
\shape italic
monotone
\shape default
sequences.
\layout Definition*
A sequence
\begin_inset Formula \( \{a_{n}\} \)
\end_inset
is
\shape italic
increasing
\shape default
if
\begin_inset Formula \( a_{n+1}\geq a_{n} \)
\end_inset
, for all
\begin_inset Formula \( n \)
\end_inset
, and is
\shape italic
decreasing
\shape default
if
\begin_inset Formula \( a_{n+1}\leq a_{n} \)
\end_inset
, for all
\begin_inset Formula \( n \)
\end_inset
.
If it is one or the other, but not jumping back and forth, then it is
\shape italic
monotone
\shape default
.
So, an increasing sequence is monotone, and a decreasing sequence is monotone.
But,
\begin_inset Formula \( a_{n}:=\sin (n) \)
\end_inset
is
\series bold
not
\series default
monotone.
\newline
A sequence
\begin_inset Formula \( \{a_{n}\} \)
\end_inset
is
\shape italic
bounded above
\shape default
if there is a number
\begin_inset Formula \( M \)
\end_inset
so that
\begin_inset Formula \( a_{n}\leq M \)
\end_inset
for all
\begin_inset Formula \( n \)
\end_inset
.
Similarly,
\begin_inset Formula \( \{a_{n}\} \)
\end_inset
is
\shape italic
bounded below
\shape default
if there is a number
\begin_inset Formula \( m \)
\end_inset
so that
\begin_inset Formula \( a_{n}\geq m \)
\end_inset
for all
\begin_inset Formula \( n \)
\end_inset
.
\begin_inset Formula \( \{a_{n}\} \)
\end_inset
is
\shape italic
bounded
\shape default
if it is bounded on both sides.
\layout Theorem
A bounded, monotonic sequence is convergent.
\layout Standard
This is a very useful result.
We'll talk about the proof a bit.
\layout Theorem
If
\begin_inset Formula \( \lim _{n\rightarrow \infty }a_{n}=A \)
\end_inset
and
\begin_inset Formula \( \lim _{n\rightarrow \infty }b_{n}=B \)
\end_inset
, then:
\begin_deeper
\layout Enumerate
\begin_inset Formula \( \lim _{n\rightarrow \infty }(a_{n}+b_{n})=A+B \)
\end_inset
.
\layout Enumerate
\begin_inset Formula \( \lim _{n\rightarrow \infty }(a_{n}-b_{n})=A-B \)
\end_inset
.
\layout Enumerate
\begin_inset Formula \( \lim _{n\rightarrow \infty }(a_{n}\, b_{n})=A\, B \)
\end_inset
.
\layout Enumerate
\begin_inset Formula \( \lim _{n\rightarrow \infty }(a_{n}/b_{n})=A/B \)
\end_inset
, if
\begin_inset Formula \( B\neq 0 \)
\end_inset
.
\layout Enumerate
\begin_inset Formula \( \lim _{n\rightarrow \infty }(c\, a_{n})=c\, A \)
\end_inset
.
\end_deeper
\layout Standard
The proof of this is not difficult, nor is it interesting.
\layout Section*
\latex latex
\backslash
S
\latex default
10.2, Series
\layout Standard
An
\shape italic
infinite series
\shape default
is an infinite sum, as we talked about before.
An unusual example of a series is given by:
\begin_inset Formula
\[
\sum _{n=0}^{\infty }b_{n},\]
\end_inset
where
\begin_inset Formula \( b_{n}=\frac{1}{n\cdot (n+1)} \)
\end_inset
, which is the ``telescoping'' series.
It converges, to 1, by the argument presented in last time's notes.
\layout Standard
The most important example of a series that
\shape italic
diverges
\shape default
is the harmonic series:
\begin_inset Formula
\[
\sum _{n=0}^{\infty }\frac{1}{n}.\]
\end_inset
This grows to
\begin_inset Formula \( \infty \)
\end_inset
as
\begin_inset Formula \( n\rightarrow \infty \)
\end_inset
.
The primary concern of this chapter will be determining whether or not
a given series will
\shape italic
converge
\shape default
.
A series
\begin_inset Formula \( \sum _{n=0}^{\infty }a_{n} \)
\end_inset
\shape italic
converges
\shape default
, to
\begin_inset Formula \( L \)
\end_inset
if
\begin_inset Formula
\[
\lim _{N\rightarrow \infty }\sum _{n=0}^{N}a_{n}=L,\]
\end_inset
in particular, the limit has to exist.
If the limit doesn't exist, the series
\shape italic
diverges
\shape default
.
\layout Standard
I've slipped something in here.
Convergence of the series
\begin_inset Formula \( \sum _{n=0}^{\infty }a_{n} \)
\end_inset
is expressed as a limit existing.
That limit is a limit of a
\shape italic
sequence
\shape default
.
Hmm.
The sequence
\begin_inset Formula \( \{S_{n}\} \)
\end_inset
whose
\begin_inset Formula \( n^{th} \)
\end_inset
term is:
\begin_inset Formula
\[
S_{n}:=\sum _{i=0}^{n}a_{i}\]
\end_inset
is called the
\shape italic
sequence of partial sums
\shape default
.
I guess that name makes sense.
\begin_inset Formula \( S_{n} \)
\end_inset
is the
\begin_inset Formula \( n^{th} \)
\end_inset
\shape italic
partial sum
\shape default
of the original series.
Since it's supposed to be easy to see whether a sequence converges, then
we talk about convergence of series in terms of the ``easier'' concept.
\layout Standard
Something else I slipped in just now is a switch in indices of summation
(from
\begin_inset Formula \( n \)
\end_inset
to
\begin_inset Formula \( i \)
\end_inset
).
We treat those just like dummy variables of integration, and call them
whatever we like, so long as we use different names for each thing that
is really different.
Just as we didn't write
\begin_inset Formula \( \int _{0}^{x}f(x)dx \)
\end_inset
in the FTC, we don't write
\begin_inset Formula \( \sum _{n=0}^{n}a_{n} \)
\end_inset
here.
\layout Standard
Here is, as I mentioned earlier, the most important series, called the
\shape italic
geometric series
\shape default
.
\layout Definition*
The series
\begin_inset Formula
\[
1+r+r^{2}+\dots =\sum _{n=0}^{\infty }r^{n}\]
\end_inset
is called the
\shape italic
geometric series
\shape default
, with
\shape italic
ratio
\shape default
\begin_inset Formula \( r \)
\end_inset
.
\layout Theorem
If
\begin_inset Formula \( -1<r<1 \)
\end_inset
, the geometric series
\begin_inset Formula
\[
1+r+r^{2}+\dots =\sum _{n=0}^{\infty }r^{n}\]
\end_inset
converges to
\begin_inset Formula \( 1/(1-r) \)
\end_inset
.
\layout Proof
By factoring the polynomial
\begin_inset Formula \( 1-r^{n+1}=(1-r)(1+r+\dots +r^{n}) \)
\end_inset
, we see that
\begin_inset Formula
\[
\frac{1}{1-r}=\sum _{i=1}^{n}r^{n}+\frac{r^{n+1}}{1-r}.\]
\end_inset
If
\begin_inset Formula \( |r|<1 \)
\end_inset
, Theorem
\begin_inset LatexCommand \ref{r^n}
\end_inset
of
\latex latex
\backslash
S
\latex default
10.1 showed that
\begin_inset Formula \( \lim _{n\rightarrow \infty }r^{n}=0 \)
\end_inset
, so the
\shape italic
remainder term
\shape default
on the right hand side will go to 0 if
\begin_inset Formula \( |r|<1 \)
\end_inset
.
Thus, for such
\begin_inset Formula \( r \)
\end_inset
, the sum on the right will converge, to
\begin_inset Formula \( 1/(1-r) \)
\end_inset
.
If
\begin_inset Formula \( |r|\geq 1 \)
\end_inset
the remainder term does not go to 0 (at 1 none of this makes sense), so
the series diverges.
\layout Standard
Now, how do you see whether or not a series converges?
\layout Theorem
\series bold
\shape up
[Rabbit Test]
\series default
\shape default
If
\begin_inset Formula \( \sum _{n=0}a_{n} \)
\end_inset
converges, then
\begin_inset Formula \( \lim _{n\rightarrow \infty }a_{n}=0. \)
\end_inset
\layout Proof
How do you show this? Well, certainly, if the terms inside the sum don't
get ever-smaller, you're going to be adding up too much.
More rigorously, if
\begin_inset Formula \( \sum _{n=0}a_{n} \)
\end_inset
converges, to
\begin_inset Formula \( L \)
\end_inset
, then you can make
\begin_inset Formula \( \sum _{i=0}^{n}a_{i} \)
\end_inset
as close as you want to
\begin_inset Formula \( L \)
\end_inset
just by taking
\begin_inset Formula \( n \)
\end_inset
large enough.
But, that means that both
\begin_inset Formula \( \sum _{i=0}^{n}a_{i} \)
\end_inset
and
\begin_inset Formula \( \sum _{i=0}^{n+1}a_{i} \)
\end_inset
are close to L, so their difference
\begin_inset Formula
\[
\left| \sum _{i=0}^{n+1}a_{i}-\sum _{i=0}^{n}a_{i}\right| =|a_{n+1}|\]
\end_inset
is as close as that to 0.
Demand it be closer, take
\begin_inset Formula \( n \)
\end_inset
larger.
Thus,
\begin_inset Formula \( \lim _{n\rightarrow \infty }a_{n}=0 \)
\end_inset
.
\layout Standard
The authors of your text call this the
\shape italic
test for divergence
\shape default
.
Whatever.
One of the professors here calls it the ``misunderstood theorem'', since
people tend to try to use it to show that a series converges
\shape italic
because
\shape default
the terms go to 0.
But that is not enough.
Necessary, but not enough.
\layout Standard
\noindent
\series bold
Examples:
\series default
\layout Enumerate
Show that
\begin_inset Formula \( \sum _{n=1}(1-1/n)^{n} \)
\end_inset
diverges.
\layout Enumerate
Find a series satisfying
\begin_inset Formula \( \lim _{n\rightarrow \infty }a_{n}=0 \)
\end_inset
but which does
\series bold
not
\series default
converge.
\layout Standard
\cursor 551
As almost an obvious thing, let me point out that if you have a convergent
series
\begin_inset Formula \( \sum _{n=0}a_{n} \)
\end_inset
, then so is
\begin_inset Formula \( \sum _{n=0}ca_{n} \)
\end_inset
for any
\begin_inset Formula \( c \)
\end_inset
, and if
\begin_inset Formula \( \sum _{n=0}b_{n} \)
\end_inset
also converges, so does
\begin_inset Formula \( \sum _{n=0}a_{n}+b_{n} \)
\end_inset
, and to the sum of the two separate ones.
Actually, I shouldn't be too cavalier about this, there are some weird
facts about convergence we might get into later that make you wonder about
this.
Besides, you should notice that I did not say anything about the series
of products
\begin_inset Formula \( \sum _{n=0}a_{n}\, b_{n} \)
\end_inset
.
That does not converge to the product of the two separate series.
There is a formula about this sort of thing, but it is more complicated
than I want to deal with right now.
\the_end
