Alternating Series, Absolute and Conditional Convergence

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ALTERNATING SERIES
1
1 1 1 1 1
    
2 3 4 5 6
series with positive terms
series with some positive
and some negative terms
1 1 1 1 1 1
1      
2 3 4 5 6 7
1 1 1 1 1
1     
2 3 4 5 6
( 1)
n 1
 1 n  odd

 1 n  even


n 1
( 1)
n 1
alternating series
n
n-th term of the series
an  (1)n  1un
un
are positive
ALTERNATING SERIES
alternating series
(1) n 1
1 1 1 1 1
 1     

n
2 3 4 5 6
n 1

alternating harmonic series

1 n
1 1 1 1
( )      

2
2 4 8 6
n 1
(1) n1
1
1
1
1

1






p
p
p
p
p
n
2
3
4
5
n 1
alternating geomtric series

alternating p-series
ALTERNATING SERIES
THEOREM: (THE ALTERNATING SERIES TEST)
1) un  0
2 ) un1  un
un  0
3) lim
n

 (1)
n 1
alternating
n 1
un

 (1)
decreasing
n 1
lim = 0
Remark:
The convergence tests that we have
looked at so far apply only to series
with positive terms. In this section and
the next we learn how to deal with
series whose terms are not
necessarily positive. Of particular
importance are alternating series,
whose terms alternate in sign.
n 1
un
convg
Example:
Determine whether the series
converges or diverges.
(1) n1

n
n 1

ALTERNATING SERIES
THEOREM: (THE ALTERNATING SERIES TEST)
1) un  0
alternating
2 ) un1  un
decreasing
un  0
3) lim
n

 (1)
n 1

Example:
Determine whether the series
converges or diverges.

2
n
n 1
(

1
)

3
n
1
n 1
un
 (1)
n 1
lim = 0
n 1
n 1
un
convg
ALTERNATING SERIES
THEOREM: (THE ALTERNATING SERIES TEST)
1) un  0
alternating
2 ) un1  un
decreasing
un  0
3) lim
n

 (1)
n 1

un
 (1)
n 1
lim = 0
n 1
n 1
un
convg
Example:
Determine whether the series
converges or diverges.

3n
( 1)

4n  1
n 1
n
ALTERNATING SERIES

L   ( 1) ui
i 1
i 1
n
S n   (1) i 1 ui
i 1
THEOREM: (THE ALTERNATING
SERIES ESTIMATION THEOREM)

satisfies the three
(1) n 1 u n
conditions
n 1

Sn
approximates the sum L of the
series with an error whose absolute
value is less than the absolute value
of the first unused term u n 1
the sum L lies between any two
successive partial sums S n and
S n 1
the remainder, Rn  L  S n has
the same sign as the first
unused term.
THEOREM: (THE ALTERNATING
Example:
SERIES ESTIMATION THEOREM)

 (1)
n 1
n 1
un
satisfies the three
conditions

 (1)
n 1
n 1
1
 0.66 6
2 n 1
1
S5  1  12  14  18  16
 0.6875
1
S6  S5  32
 0.65625
L  Sn  un1
L  S5 
1
32
S n  L  S n1
OR
0.65625
 S6  L  S5  0.6875
S n1  L  S n
sign ( L  S n ) 
 sign (1st unused)
sign ( L  S5 )  negative
ALTERNATING SERIES

n
S n   (1) i 1 ui
L   ( 1) ui
i 1
i 1
i 1
THEOREM: (THE ALTERNATING
SERIES ESTIMATION THEOREM)

 (1)
n 1
n 1
un
satisfies the three
conditions
L  Sn  un1
S n  L  S n1
OR
S n1  L  Sn
sign ( L  S n ) 
 sign (1st unused)
Example:
Find the sum of the series
correct to three decimal places.
(1) n1

n!
n 0

ALTERNATING SERIES
Example:
Find the sum of the series correct
to three decimal places.
(1) n1

n!
n 0

ALTERNATING SERIES

n
S n   (1) i 1 ui
L   ( 1) ui
i 1
i 1
i 1
THEOREM: (THE ALTERNATING
SERIES ESTIMATION THEOREM)

 (1)
n 1
n 1
satisfies the three
u n conditions
L  Sn  un1
S n  L  S n1
OR
S n1  L  Sn
sign ( L  S n ) 
 sign (1st unused)
REMARK:
The rule that the error is smaller than the
first unused term is, in general, valid only
for alternating series that satisfy the
conditions of the Alternating Series
Estimation Theorem. The rule does not
apply to other types of series.
ALTERNATING SERIES
TERM-102
ALTERNATING SERIES
TERM-101
ALTERNATING SERIES
TERM-092
Alternating Series, Absolute and Conditional Convergence

 an  1 
n 1

a
n 1
n
 1
1 1 1 1 1
    
2 3 4 5 6

a
n 1
1
1
1
1
1





2 2 32 4 2 5 2 6 2


n 1
(1) n 1
1 1 1 1 1

1

    

n
2 3 4 5 6
n 1

DEF:

a
n 1
Is called
Absolutely convergent
converges absolutely
IF
 1
an  1 
1 1 1 1 1
    
2 3 4 5 6
1
1
1
1
1





2 2 32 4 2 5 2 6 2

1
1 1 1 1 1

1

    

n
2
3 4 5 6
n 1
Example:

n
n
a
n 1
n
convergent
Test the series for absolute
convergence.
(1) n1

2
n
n 1

Alternating Series, Absolute and Conditional Convergence
DEF:

a
n 1
Example:

n
Is called
Absolutely convergent
a
IF
n 1
n
convergent
converges absolutely
DEF:
n 1

sin( n)

2
n
n 1
Example:

a
Test the series for absolute
convergence.
n
Is called conditionally convergent
Test the series for absolute
convergence.
if it is convergent but not absolutely convergent.
REM:

a
n 1
convg
n
a
n 1
(1) n1

n
n 1


n
divg
Alternating Series, Absolute and Conditional Convergence
DEF:

a
n 1
Example:

n
Is called
Absolutely convergent
a
IF
n 1
n
convergent
converges absolutely
DEF:

a
n 1
n
Is called conditionally convergent
if it is convergent but not absolutely convergent.
REM:

a
n 1
convg

n
a
n 1
n
divg
Test the series for absolute
convergence.
Alternating Series, Absolute and Conditional Convergence

THM:
a
n 1
n
Absolutely
convergent

THM:
a
n 1

a
n 1
n
convergent
n
convg

convg
n
a
n 1
Example:
Determine whether the series
converges or diverges.
cosn 

2
n
n 1

Alternating Series, Absolute and Conditional Convergence

a
n 1

 an
n 1

a
n 1
n
conditionally
convergent
Absolutely
convergent

a
n 1
n
convergent
n
divergent

a
n 1

n
a
n 1
n
Alternating Series, Absolute and Conditional Convergence
DEF:

a
n 1
Example:

n
Is called
Absolutely convergent
a
IF
n 1
n
convergent
converges absolutely
DEF:

a
n 1
n
Is called conditionally convergent
if it is convergent but not absolutely convergent.
REM:

a
n 1
convg

n
a
n 1
n
divg
Choose one: absolutely
convergent or conditionally
convergent
Alternating Series, Absolute and Conditional Convergence
REARRANGEMENTS

n
(

1
)
Divergent

n 1
 1  1  1  1  1  1  1  1  1  
 1  (1  1)  (1  1)  (1  1)  (1  1)    1
 (1  1)  (1  1)  (1  1)  (1  1)    0
If we rearrange the order of the terms in a finite sum, then of
course the value of the sum remains unchanged.
But this is not always the case for an infinite series.
By a rearrangement of an infinite series we mean a series
obtained by simply changing the order of the terms.
Alternating Series, Absolute and Conditional Convergence
REARRANGEMENTS

n
(

1
)
Divergent

n 1
 1  1  1  1  1  1  1  1  1  
 1  (1  1)  (1  1)  (1  1)  (1  1)    1
 (1  1)  (1  1)  (1  1)  (1  1)    0
(1) n1

n
n 1

convergent
See page 719
 1  12  13  14  15  16  17 
 1  12  13  14  15  16  17   ln 2
 1  12  13  14  15  16  17   12 ln 2
Alternating Series, Absolute and Conditional Convergence
REARRANGEMENTS
REMARK:

 an
n 1
Absolutely
convergent
any rearrangement
has the same sum s
with sum s
Riemann proved that

 an
n 1
Conditionally
convergent
r is any real number
there is a rearrangement
that has a sum equal to r.
SUMMARY OF TESTS
SUMMARY OF TESTS
Special Series:
Series Tests

lim an  0
1) Test for Divergence
2) Integral Test


1
n 
2) Harmonic Series
f ( x)dx
bn  an
n
an
7) Alternating Series Test
4) p-series

alt , dec, lim 0
1
n

 (b b
1

p
n 1 n
5) Alternating p-series
an 1
n  a
n
L  lim
n 

3) Telescoping Series
bn
L  lim
n 1
n 1
4) Limit Comparison Test c  lim an
n 
6) Root Test
1) Geometric Series
n 1

3) Comparison Test
5) Ratio Test
 ar
n
n 1
n 1
)
(1) n

np
n 1

SUMMARY OF TESTS
1) Determine whether
convg or divg
2) Find the sum s
5-types
3) Estimate the
sum s
5) Partial
sums
4) How many
terms are needed
within error
ALTERNATING SERIES
TERM-101
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