Download Part 6/Series of constants

Math 104
Calculus I
Part 6
INFINITE SERIES
Series of Constants
We’ve looked at limits and sequences.
Now, we look at a specific kind of
sequential limit, namely the limit (or
sum) of a series.
Zeno’s Paradox
How can an infinite number of things happen
in a finite amount of time?
(Zeno's paradox concerned Achilles and a
tortoise)
Discussion Questions
1. Is Meg Ryan’s reasoning correct? If it isn't,
what is wrong with it?
2. If the ball bounces an infinite number of
times, how come it stops? How do you
figure out the total distance traveled by the
ball?
Resolution
The resolution of these problems is
accomplished by the use of limits.
In particular, each is resolved by
understanding why it is possible to "add
together" an infinite number of numbers and
get a finite sum.
An example
Meg Ryan worried about adding together
1 1 1 1
    ...
2 4 8 16
Picture This
The picture suggests that
the "infinite sum"
1 1 1 1
    ...
2 4 8 16
should be 1. This is in fact true, but requires
some proof.
We'll provide the proof, but in a more general
context.
The idea of a series
A "series" is any "infinite sum" of numbers.
Usually there is some pattern to the numbers, so
we can give an idea of the pattern either by giving
the first few numbers, or by giving an actual
formula for the nth number in the list. For
example, we could write
n


1 1 1 1
1
1
    ... as  n or as   
2 4 8 16
n 1 2
n 1  2 
The things being added together are called “terms” of
the series.
Other series we will consider...

1 1 1
1
1     ... or 
2 3 4
n 1 n
this is sometimes called the " harmonic series"

 1
1 1 1
1     ... or 
2 3 4
n
n 1
n 1
this is called the " alternatin g harmonic series"
1 1 1
1
1
  
  ..., which you could recognize as
2 6 12 20 30

1

n 1 nn  1

1 1 1 1
1
1      ..., or  (since 0! is defined to equal 1).
1 2! 3! 4!
n  0 n!
Two obvious questions
1. Does the series have a sum? (Officially:
"Does the series converge?")
2. What is the sum? (Officially: "What does
the series converge to?")
A less obvious question is...
3. How fast does the series converge?
Convergence
The word convergence suggests a limiting
process. Fortunately, we don't have to
invent a new kind of limit for series.
Think of series as a process of adding together
the terms starting from the beginning. Then
the nth "partial sum" of the series is simply
the sum of the first n terms of the series.
For example...
the partial sums of the IQ series are:
1st partial sum = 1/2
2nd partial sum = 1/2 + 1/4 = 3/4
3rd partial sum = 1/2 + 1/4 + 1/8 = 7/8
and so forth.
It looks line the
nth
partial
sum
of
the
IQ
n
2
1
series is
n
2
It is only natural
It is natural to define (and this is even the
official definition!) the sum or limit of the
series to be equal to the limit of the sequence
of its partial sums, if the latter limit exists.
For the IQ series, we really do have:
n
2

1
lim sn  lim
1
n 
n 
n
2
This bears out our earlier suspicion.
This presents a problem...
The problem is that it is often difficult or
impossible to get an explicit expression for
the partial sums of a series.
So, as with integrals, we'll learn a few basic
examples, and then do the best we can -sometimes only answering question 1, other
times managing 1 and 2, and still other
times 1, 2, and 3.
Geometric series
The IQ series is a specific example of a
geometric series .
A geometric series has terms that are (possibly
a constant times) the successive powers of a
number.
The IQ series has successive powers of 1/2.

1  1  1  1  ...1n
Other examples
n 1

3
3
3
3
1
0.3333333...  


 ... 3 
10 100 1000 10000
n 1  10 

3  12  48  192  ...   3(4 n )
n 0
5 5
5
 1 
5  
 ...   5 
7 49 343
n 0  7 


3 3
3
3
3
 

 ...   n
32 64 128 256
n 5 2
n
n
Convergence of geometric series
Start (how else?) with partial sums
Finite geometric sum:
a  ar  ar 2  ar 3  ...  ar n  S n
ar  ar 2  ar 3  ar 4  ...  ar ( n1)  rS n
Therefore
( n 1)

(1  r ) S n  a(1  r
( n 1)
a
(
1

r
)
and so
Sn 
1 r
We conclude that...
a
lim S n is equal to
if r  1, and does not exist otherwise.
n 
1 r
Therefore the geometric series converges precisely when
r  1, and diverges otherwise.
connect
Some questions
Which of the geometric series on the previous
slide (reproduced on the next slide)
converge?
What do they converge to?

1  1  1  1  ...1n
Other examples
n 1

3
3
3
3
1
0.3333333...  


 ... 3 
10 100 1000 10000
n 1  10 

3  12  48  192  ...   3(4 n )
n 0
5 5
5
 1 
5  
 ...   5 
7 49 343
n 0  7 


3 3
3
3
3
 

 ...   n
32 64 128 256
n 5 2
n
n
Telescoping series
Another kind of series that we can sum:
telescoping series
This seems silly at first, but it's not!
A series is said to telescope if all the terms in
the partial sums cancel except perhaps for
the first and the last.
Example
1
1 1
1 1
1
1
(1  )  (  )  (  )  ...( 
)  ...
2
2 3
3 4
n n 1
1
Clearly th e nth partial sum of this series is 1 ,
n 1
so the series converges to 1.
What’s the big deal?
Well, you could rewrite the series as
1 1 1
1
   ... 
 ...
2 6 12
n(n  1)
which is not so obvious (in fact, it was one
of the examples given near the beginning of
today’s class).
Now you try one...

1
What is the sum of the series  2 ?
k 2 k  1
A)
B)
C)
D)
E)
1
3/4
1/2
1/4
1/8
Improper integrals
Occasionally it helps to recognize a series as a
telescoping series. One important example of such a
series is provided by improper integrals.
Suppose F '(x) = f(x). Then we can think of the

improper integral  f ( x)dx
1
as being
the
sum
of
the
series
2
3
4
 f ( x)dx   f ( x)dx   f ( x)dx  ... 
1
2
3
(F(2) - F(1))  (F(3) - F(2))  (F(4) - F(3))  ...
Continued...
Since the nth partial sum of this series is F(n+1) F(1), it's clear that the series converges to
(lim F(n))  F(1)
n 
be equal to
just as the integral would
(lim F( x))  F(1)
x 
(Note the subtle difference between the two limits
-- the limit of the series might exist even when the
improper integral does not).
The convergence question
For a while, we’ll concentrate on the question
1: Does the series converge?
One obvious property that convergent series
must have is that their terms must get
smaller and smaller in order for the limit of
the partial sums to exist.
Fundamental necessary
condition for convergence:

A series
a
n 1
n
cannot converge unless lim an  0
n 
This is only a test you can use
to prove that a series does
NOT
converge


(e.g.,  nn1 diverges, as does  arctan( n))
n 1
n 1
Harmonic
Just because the nth term goes to zero doesn't mean that
the series converges. An important example is the
harmonic series


n 1
1
n
We can show that the harmonic
series diverges by the following
argument using the partial sums:
For the harmonic series,
S1  1
S 2  1  12 
3
2
S 4  1  12  13  14  32  14  14 
4
2
S8  1  12  13  14  15  16  17  18  42  18  18  18  18 
5
2
Harmonic (cont.)
and so on -- every time we double the number
of terms, we add at least one more half. This
indicates (and by induction we could prove)
that
S 2n 
n 3
2
, so it is necessaril y the case that
lim S n  , so the harmonic series diverges.
n 
Cantilever tower:
The divergence of the harmonic
series makes the following trick
possible. It is possible to stack
books (or cards, or any other
kind of stackable, identical
objects) near the edge of a table
so that the top object is
completely off the table (and as
far off as one wishes, provided
you have enough objects to
stack).
Series of positive terms
Convergence questions for series of positive terms
are easiest to understand conceptually.
Since all the terms an are assumed to be positive, the
sequence of partial sums {S n} must be an
increasing sequence.
So the least upper bound property discussed earlier
comes into play -- either the sequence of partial
sums has an upper bound or it doesn't.
If the sequence of partial sums is bounded above,
then it must converge and so will the series. If not,
then the series diverges. That's it.
Tests for convergence of series
of positive terms:
The upper bound observations give rise to
several "tests" for convergence of series of
positive terms. They all are based pretty much
on common sense ways to show that the partial
sums of the series being tested is bounded are
all less than those of a series that is known to
converge (or greater than those of a series that
is known to diverge). The names of the tests we
will discuss are...
Tests...
TODAY
1. The integral test
TODAY
2. The comparison test
3. The ratio test
4. The limit comparison test (sometimes
called the ratio comparison test)
5. The root test
The integral test
Since improper integrals of the form

1
f ( x )dx
provide us with many examples of telescoping
series whose convergence is readily determined,
we can use integrals to determine convergence
of series:

For example, consider the series

n 1
1
n2
From the following picture, it is evident that
the nth partial sum of this series is less than
n
1  1
1
x2
dx
Integral test cont.
The sum of the terms is equal to the sum of the areas
What
is the
of the shaded
rectangles,
and sum?
if we start
integrating at 1 instead of 0, the
improper integral


1
1 x2
dx converges
(question: what is the integral? so what bound to
you conclude for the series?).
Since the value of the improper integral (plus 1)
provides us with an upper bound for all of the
partial sums, the series must converge.
It is an interesting question as to exactly what the
sum is. We will answer it next week.
The integral test...
says that if the function f(x) is positive and

decreasing, the series
 f ( x ) and the improper
x 1

1
integral f ( x )dx either both converge or both
diverge (only convergence at infinity needs to
be checked for the integral). This gives us a new,
easier proof of the divergence of the harmonic
series - -because we already knew the divergence of
the integral


1 dx.
1 x
Discussion and Connect
Question…
-- for which exponents p does the series


n 1
1
np
converge?
(These are sometimes called p-series, for obvious
reasons -- these together with the geometric series
give us lots of useful examples of series whose
convergence or divergence we know).
Error estimates:
Using the picture that proves the integral test for
convergent series, we can get an estimate on how far
off we are from the limit of the series if we stop
adding after N terms for any finite valueof N.
If we approximate the convergent series
by the partial sum s N 
N
 f ( n)
 f (n)
n 1
n 1
then the error we commit is less than the value of the

integral
f ( x)dx

N
Take a closer look...
For example, the sum 1  14  19  161  251  5269
3600 , which
is approximat ely 1.46. This differs by less than


1
2
5 x

dx  15 , or 0.2, from the infitie sum  n12 . (As
n 1
we shall see laer, this estimate isn' t far off - - the
actual sum is a little bigger tha n 1.6
Question

Does the series

n 1
A) Converge
B) Diverge
n
1 n 2
converge or diverge?
Question

Does the series

n 1
A) Converge
B) Diverge
arctan( n )
1 n
2
converge or diverge?
Exercise
Connect
For this latter series, find a bound on the
error if we use the sum of the first 100
terms to approximate the limit. (answer:
it is less than about .015657444)
The comparison test
This convergence test is even more commonsensical than the integral test. It says that if

all the terms of the series
a
n 1
are less than
n

the corresponding terms of the series

and if
b
n 1
n
n 1

converges, then
converges also.
a
n 1
b
n
n
Reverse
This test can also be used in reversed -- if
the b series diverges and the a’s are bigger
than the corresponding b’s, then  an
n 1
diverges also.
Examples:


n 1
1
2n  n
converges.


n 1
n
n  sin(n )
diverges.
Question

Does the series

k 5
A) Converge
B) Diverge
1
k 2
converge or diverge?
Question

Does the series

n 1
A) Converge
B) Diverge
1
2n  n 2
converge or diverge?