Download infinite series

Many quantities that arise in applications cannot be computed exactly.
We cannot write down an exact decimal expression for the number π
or for values of the sine function such as sin(1). However, sometimes
these quantities can be represented as infinite sums. For example,
using Taylor series, we can show that
1 1 1 1 1
sin 1  Tn 1  1     
 
3! 5! 7! 9! 11!
This is an example of an infinite series.
but isn't sin 1 a finite number?
x3 x5 x 7
McLaurin Series for f  x   sin x  Tn  x   x     
3! 5! 7!
1 1 1 1 1
   
 
3! 5! 7! 9! 11!
What precisely does this mean? It’s impossible to add up infinitely
many numbers, but what we can do is compute the partial sums SN,
defined as the finite sum of the terms up to and including Nth term.
Here are the first five partial sums of the infinite series for sin(1):
sin 1  1 
S1  1
Compare these values with
the value obtained from a
calculator:
1
S 2  1   0.833
3!
1 1
S3  1    0.841667
3! 5!
1 1 1
S 4  1     0.841468
3! 5! 7!
1 1 1 1
S5  1      0.8414709846
3! 5! 7! 9!
We will eventually prove that sin 1  lim S N :
N 
(Example 2 in Section 11.7). So although we cannot add up infinitely
many numbers, it makes sense to define the sum of an infinite series as
a limit of partial sums.
In general, an infinite series is an expression of the form

a
n 1
n
 a1  a2  a3  a4  
where {an} is any sequence.
N
S N   an  a1  a2  a3    aN
n 1
The Nth partial sum SN is the finite sum of the terms up to and
including aN:
DEFINITION Convergence of an Infinite Series An infinite series

a
nk
n
converges to the sum S, if its partial sums converge to S:

lim S N  S   an
N 
nk
• If the limit does not exist, we say that the infinite series diverges.
• If the limit is infinite, we say that the infinite series diverges to
infinity.
Telescoping Series Investigate numerically:

1
1
1
1
1




 
1 2  2  3  3  4  4  5 
n 1 n  n  1
S 
Numerically :
The values of the partial sums suggest convergence to S = 1. To prove
this, we observe that because of the identity, each partial sum collapses
down to just two terms:
After cancellation, a series whose partial
sums only have a fixed number of terms.
Now compute the sum S by rewriting the summand using partial
fractions:
1
A
B
 
 1  A  n  1  Bn
n  n  1 n n  1
1
1
1
n  1  B  1; n  0  A  1 
 
n  n  1 n n  1
Telescoping Series Investigate numerically:

1
1
1
1
1
1 1 S




 

S1  
1 2  2  3  3  4  4  5 
n 1 n  n  1
1 2
1
1
1 1   1 1 
 SN  1 
S2          1 
N 1
3
1 2   2 3 
QED
1 

1
 1 1 1
 S  lim S N  lim 1 
 1
S3   1        1 
N 
N 
 N 1 
4
 3 3 4
The values of the partial sums suggest convergence to S = 1. To prove
this, we observe that because of the identity, each partial sum collapses
down to just two terms:
A series whose partial sums only have a
fixed number of terms after cancellation.
Now compute the sum S by rewriting the summand using partial
fractions:
1
A
B
 
 1  A  n  1  Bn
n  n  1 n n  1
1
1
1
n  1  B  1; n  0  A  1 
 
n  n  1 n n  1
In most cases (apart from telescoping series and the geometric series
introduced below), there is no simple formula for the partial sum SN.
Therefore, we shall develop techniques that do not rely on formulas for
SN.
an  a1 , a2 , a3 , a4 , ...
It is important to keep in mind the difference between a sequence {an}
and an infinite series

a .
n 1
n
Sequences versus Series Discuss the difference between {an} and

1
an , where an 
.

n  n  1
n 1
1
1
1
,
,
, ....
The sequence is the list of numbers
1 2  2  3  3  4 
1
0
n  n  n  1
lim an  lim
This sequence converges to…
n 
The infinite series is the sum of the numbers an, defined formally as the
limit of the partial sums. This sum is not zero. In fact, the sum is equal
to 1 by Example 1:


1
1
1
1
an  



   1

1 2  2  3  3  4 
n 1
n 1 n  n  1
• With a sequence, we consider the limit of the individual terms an.
• With a series, we are interested in the sum of the terms
a1 + a2 + a3 +… which is defined as the limit of the partial sums.
THEOREM 1 Linearity of Infinite Series
a
n
and
b
n
converge    an  bn  and
 ca
converge (c is any constant), and
 a  b  a  b 
 a  b  a  b 
 ca  c a ,  c is any constant 
n
n
n
n
n
n
n
n
n
n
n
also
A main goal in this chapter is to develop techniques for determining
whether a series converges or diverges. It is easy to give examples of
series that diverge:

•
S  1
diverges to infinity (the partial sums increase without bound):
n 1

•
S1 = 1, S2 = 2, S3 = 3, S4 = 4,…
S    1
n 1
diverges (the partial sums jump between 1 and 0):
n 1
S1 = 1, S2 = 0, S3 = 1, S4 = 0,…
The terms oscilate between 1 &  1, but...
Next, we study the geometric series, which converge or diverge
depending on the common ratio r.
Geometric Series:

S   cr n  c  cr  cr 2  cr 3  cr 4  cr 5  
n 0
For r  1/ 2 and c = 1, we can visualize the geometric series
starting at n = 1:
Adding up the terms corresponds to
moving stepwise from 0 to 1, where each
step is a move to the right by half of the
remaining distance. Thus S = 1.

1 1 1 1 1
S   n        1
2 4 8 16
n 1 2
There is a simple formula for computing the partial sums of a geometric
n0
series:
2
3
N
S N  c  cr  cr  cr    cr
rS N  cr  cr 2  cr 3    cr N  cr N 1
S N  rS N  c  cr N 1
S N 1  r   c 1  r N 1 
S N  c  cr  cr 2  cr 3     cr N 
c 1  r N 1 
1 r
This formula enables us to sum the geometric series.
Multiply the series by the common ratio  r  .
The difference of S N & rS N
Solve for S N
Factor
THEOREM 2 Sum of a Geometric Series Let c  0.
If |r| < 1, then

0
c
r
n
2
3
cr

c

cr

cr

cr
  

1 r
n 0

M
cr
n
M
M 1
M 2
M 3
cr

cr

cr

cr

cr
  

1 r
nM
If |r| ≥ 1, then the geometric series diverges.
S N  c  cr  cr  cr    cr 
2
3
N
c 1  r
N 1
1 r

an  cr

Evaluate
n
n

5
1
1  =

4
n 0  5 
n
5
 .
n 0

c
cr  c  cr  cr  cr   

1 r
n 0
n

2
n
3
3
4
5
 3
 3
 3
 3
Evaluate  7     7     7     7     .
4
 4
 4
 4
n 3 

M
cr
n
M
M 1
M 2
M 3
cr

cr

cr

cr

cr
  

1 r
nM
3
 3
7 
n
M

cr
27
4
 3

7  



4  1 r
7/4
16
n 3 
2  3n
Evaluate S   n .
5
n 0


c
cr  c  cr  cr  cr   

1 r
n 0
n
2
3
 1 3 
1
 3
S    2        2      


5
5
5
5








n 0 
n

0
n

0

2
1
5 5


  5
4/5 2/5 2 2

n
n

n

n

Conjecture whether
  1
n 1
converges or diverges?
n 1
n
1 2 3 4
     
n 1 2 3 4 5
We'll evaluate S in a moment.
THEOREM 3 Divergence Test If the nth term {an } does not
converge to zero, then the series

a
n 1
n
diverges.
Contrapositive
The Divergence Test (also called the nth-Term Test) is often stated as follows:
LOGIC 101
Statement: P  Q

a
T
Converse: Q  P ???
Contrapositive: ~Q ~ P
T
n 1
n
converges  lim an  0
n 
The converse of a true statement, does not have to be true.

a
n 1
n
converges  lim an  0
by the contrapositive of the nth-Term Test
n 
nth-Term Test
Prove the divergence of S 

n
.

n 1 4n  1
1
lim an   S diverges.
n 
4
QED
Determine the convergence or divergence of

S    1
n 1
n 1
n
1 2 3 4
     
n 1 2 3 4 5
n
n 1
 1, &  1 diverges  S diverges
n 1
The Divergence Test tells only part of the story. If {an} does not tend to
zero, then  an certainly diverges. But what if an does tend to zero? In
this case, the series may converge or it may diverge. In other words,
lim an  0 is a necessary condition of convergence, but it is not
n 
sufficient. As we show in the next example, it is possible for a series to
diverge even though its terms tend to zero.
Sequence Tends to Zero, yet the Series Diverges Prove the
divergence of

n
S 
n 1
1
1
1
1




n
1
2
3
Converse

lim an  0   an converges
n 
n 1
1/ N  0. Because each previous term in the sum SN is greater than
1/ N , we have
QED

1
1
1
1


  
n
1
2
N
n 1
1
1
1


  
N
N
N
 1 
 N
 N
 N
SN  
This shows that S N  N . But N increases without bound.
Therefore SN also increases without bound. This proves that the series
nth-Term Test  a converges  lim a  0
diverges.

n 1
n
n 
n
chosen in a similar fashion relative to AB and E is chosen
relative to BC , then
Geometric series were used as
early as the third century BC by
Archimedes in a brilliant
argument for determining the
area S of a “parabolic segment”.
Given two points A and C on a
parabola, there is a point B
between A and C where the
tangent line is parallel to AC
(apparently, Archimedes knew
the Mean Value Theorem more
than 2000 years before the
invention of calculus). Let T be
the area of triangle ΔABC.
Archimedes proved that if D is
This construction of triangles can be continued. The next
step would be to construct the four triangles on the
of total area
segments
. Then
construct eight triangles of total area
, etc. In this way,
we obtain infinitely many triangles that completely fill up
the parabolic segment. By the formula for the sum of a
geometric series,
For this and many other achievements, Archimedes is
ranked together with Newton and Gauss as one of the
greatest scientists of all time.
The modern study of infinite series began in the
seventeenth century with Newton, Leibniz, and their
contemporaries. The divergence of
(called the harmonic series) was known to the medieval
scholar Nicole d’Oresme (1323–1382), but his proof was
lost for centuries, and the result was rediscovered on more
than one occasion. It was also known that the sum of the
reciprocal squares
converges, and in the 1640s,
the Italian Pietro Mengoli put forward the challenge of finding its sum. Despite the efforts of the
best mathematicians of the day, including Leibniz and the Bernoulli brothers Jakob and Johann,
the problem resisted solution for nearly a century. In 1735, the great master Leonhard Euler (at
the time, 28 years old) astonished his contemporaries by proving that
This formula, surprising in itself, plays a role in a variety of mathematical fields. A theorem
from number theory states that two whole numbers, chosen randomly, have no common factor
with probability 6/π2 ≈ 0.6 (the reciprocal of Euler’s result). On the other hand, Euler’s result
and its generalizations appear in the field of statistical mechanics.
Archimedes (287 BCE–212 BCE), who discovered the law of the lever, said “Give me a place to
stand on, and I can move the earth” (quoted by Pappus of Alexandria c. AD 340).
Archimedes showed that the area S
of the parabolic segment is
where T is the area of ΔABC.
Suggested Problems
1-5 odd,11,15,17,21,23,27,31,37,41,45