Transcript infinite series

9
Infinite Series
Copyright © Cengage Learning. All rights reserved.
9.2
Series and Convergence
Copyright © Cengage Learning. All rights reserved.
Objectives
 Understand the definition of a convergent
infinite series.
 Use properties of infinite geometric series.
 Use the nth-Term Test for Divergence of an
infinite series.
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Sequences
A sequence is defined as a function whose domain is the set of positive integers.
Although a sequence is a function, it is common to represent sequences by subscript
notation rather than by the standard function notation.
For instance, in the sequence
1 is mapped onto a1, 2 is mapped onto a2, and so on.
The numbers a1, a2, a3, . . ., an, . . . are the terms of the sequence.
The number an is the nth term of the sequence,
and the entire sequence is denoted by {an}.
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Limit of a Sequence
Sequences whose terms approach to limiting values,
are said to converge.
For instance, the sequence {1/2n}
converges to 0.
We are not going to spend much time on theory of
sequences because sequence is just a function, so
we can apply all the limit theory we learned in Calculus 1
Converges to L crossing it only once
Converges to L oscillating
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Infinite Series
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One important application of infinite sequences is in representing “infinite summations.”
Informally, if {an} is an infinite sequence, then
is an infinite series (or simply a series).
The numbers a1, a2, a3, are the terms of the series.
For some series it is convenient to begin the index at n = 0 (or some other integer).
As a typesetting convention, it is common to represent an infinite series as simply
In such cases, the starting value for the index must be taken from the context of the statement.
To find the sum of an infinite series, consider the following sequence of partial sums.
If this sequence of partial sums
converges, the series is said to
converge.
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n
1
1  
n
n
1 r
1 2
1
Sn  a

 1     1 as n  
1 r 2 1 1
2
2
Maxima fails
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Infinite Series
The series
is a telescoping series of the form
Note that b2 is canceled by the second term, b3 is canceled by the third term, and so on.
Because the nth partial sum of this series is
S n = b1 – bn + 1
=> telescoping series will converge if and only if bn approaches a finite number as
Moreover, if the series converges, its sum is
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Apart[ ] breaks rational
functions to partial fractions
Maxima fails
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Geometric Series
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
If r  -1   a (1) n  a  a  a  a...
n 0
a  n  odd 
Sn  
  diverges
0  n  even

If r  1   a(1) n  a  a  a  a...
n 0
 na    diverges
If | r | 1 | r |n    diverges
either goes to infinity or oscillates
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Example 3(a) – Convergent and Divergent Geometric Series
The geometric series
has a ratio of
with a = 3.
Because 0 < |r| < 1, the series converges
and its sum is
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Example 3(b) – Convergent and Divergent Geometric Series
cont’d
The geometric series
has a ratio of
Because |r| ≥ 1, the series diverges.
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My example:


1
2
1  3
1
 10
 n   5 2
  n  5 1  6  3
 2

2 
3 n 1 2
3
n 1  4n  1
n 1 4n  1
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nth-Term Test for Divergence
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The contrapositive of Theorem 9.8 provides a useful test for divergence.
This nth-Term Test for Divergence states that if the limit of the nth term of a series
does not converge to 0, the series must diverge.
My example :
 3n 2 
3n 2
3



a


 0  diverges

n
2
 4n 2  1 
4
n

1
4
n 1 


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Because the limit of the nth term is 0, the nth-Term Test for Divergence does not apply
and you can draw no conclusions about convergence or divergence.
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