11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
11.6
Absolute Convergence
and the Ratio and Root tests
In this section, we will learn about:
Absolute convergence of a series
and tests to determine it.
ABSOLUTE CONVERGENCE
Given any series Σ an, we can consider
the corresponding series

a
n 1
n
 a1  a2  a3  ...
whose terms are the absolute values of
the terms of the original series.
ABSOLUTE CONVERGENCE
Definition 1
A series Σ an is called absolutely
convergent if the series of absolute
values Σ |an| is convergent.
ABSOLUTE CONVERGENCE
Notice that, if Σ an is a series with
positive terms, then |an| = an.
 So, in this case, absolute convergence
is the same as convergence.
ABSOLUTE CONVERGENCE
The series
n 1
(1)

2
n
n 1

Example 1
1 1 1
 1  2  2  2  ...
2 3 4
is absolutely convergent because


n 1
n 1
(1)
2
n

1
1 1 1
  2  1  2  2  2  ...
2 3 4
n 1 n
is a convergent p-series (p = 2).
ABSOLUTE CONVERGENCE
Example 2
We know that the alternating harmonic series
(1)

n
n 1

n 1
1 1 1
1     ...
2 3 4
is convergent.
 See Example 1 in Section 11.5.
Example 2
ABSOLUTE CONVERGENCE
However, it is not absolutely convergent
because the corresponding series of absolute
values is:


n 1
(1)
n
n 1

1
1 1 1
 1     ...
2 3 4
n 1 n
 This is the harmonic series (p-series with p = 1)
and is, therefore, divergent.
CONDITIONAL CONVERGENCE
Definition 2
A series Σ an is called conditionally
convergent if it is convergent but not
absolutely convergent.
ABSOLUTE CONVERGENCE
Example 2 shows that the alternating
harmonic series is conditionally convergent.
 Thus, it is possible for a series to be convergent
but not absolutely convergent.
 However, the next theorem shows that absolute
convergence implies convergence.
ABSOLUTE CONVERGENCE
If a series Σ an is
absolutely convergent,
then it is convergent.
Theorem 3
ABSOLUTE CONVERGENCE
Theorem 3—Proof
Observe that the inequality
0  an  an  2 an
is true because |an| is either an or –an.
ABSOLUTE CONVERGENCE
Theorem 3—Proof
If Σ an is absolutely convergent, then Σ |an|
is convergent.
So, Σ 2|an| is convergent.
 Thus, by the Comparison Test, Σ (an + |an|)
is convergent.
ABSOLUTE CONVERGENCE
Theorem 3—Proof
Then,
a (a
n
n
 an )   an
is the difference of two convergent series
and is, therefore, convergent.
ABSOLUTE CONVERGENCE
Example 3
Determine whether the series

cos n cos1 cos 2 cos3
 2  2  2  ...

2
1
2
3
n 1 n
is convergent or divergent.
ABSOLUTE CONVERGENCE
Example 3

cos n cos1 cos 2 cos3
 2  2  2  ...

2
1
2
3
n 1 n
The series has both positive and negative
terms, but it is not alternating.
 The first term is positive.
 The next three are negative.
 The following three are positive—the signs change
irregularly.
Example 3
ABSOLUTE CONVERGENCE
We can apply the Comparison Test to
the series of absolute values:


n 1

cos n
cos n
 2
2
n
n
n 1
ABSOLUTE CONVERGENCE
Example 3
Since |cos n| ≤ 1 for all n, we have:
cos n
1
 2
2
n
n
 We know that Σ 1/n2 is convergent
(p-series with p = 2).
 Hence, Σ (cos n)/n2 is convergent
by the Comparison Test.
ABSOLUTE CONVERGENCE
Example 3
Thus, the given series Σ (cos n)/n2
is absolutely convergent and, therefore,
convergent by Theorem 3.
ABSOLUTE CONVERGENCE
The following test is very useful
in determining whether a given series
is absolutely convergent.
Case i
THE RATIO TEST
an 1
lim
 L 1
n  a
n

then the series  an is absolutely convergent
If
n 1
(and therefore convergent).
Case ii
THE RATIO TEST
If
an 1
lim
 L 1
n  a
n
or
an 1
lim

n  a
n

then the series
a
n 1
n
is divergent.
THE RATIO TEST
If
Case iii
an 1
lim
1
n  a
n
the Ratio Test is inconclusive.
 That is, no conclusion can be drawn about
the convergence or divergence of Σ an.
THE RATIO TEST
Case i—Proof
The idea is to compare the given series
with a convergent geometric series.
 Since L < 1, we can choose a number r
such that L < r < 1.
Case i—Proof
THE RATIO TEST
Since
an1
lim
L
n  a
n
and
Lr
the ratio |an+1/an| will eventually be less than r.
 That is, there exists an integer N such that:
an1
r
an
whenever n  N
THE RATIO TEST
i-Proof (Inequality 4)
Equivalently,
|an+1| < |an|r
whenever n ≥ N
THE RATIO TEST
Case i—Proof
Putting n successively equal to N, N + 1,
N + 2, . . . in Equation 4, we obtain:
|aN+1| < |aN|r
|aN+2| < |aN+1|r < |aN|r2
|aN+3| < |aN+2| < |aN|r3
THE RATIO TEST
i-Proof (Inequality 5)
In general,
|aN+k| < |aN|rk
for all k ≥ 1
Case i—Proof
THE RATIO TEST
Now, the series

a
k 1
r  aN r  aN r  aN r  ...
k
N
2
3
is convergent because it is a geometric series
with 0 < r < 1.
THE RATIO TEST
Case i—Proof
Thus, the inequality 5, together with
the Comparison Test, shows that the series


n  N 1

an   aN k  aN 1  aN  2  aN 3  ...
k 1
is also convergent.
THE RATIO TEST
It follows that the series

Case i—Proof
is convergent.
a
 n
n 1
Recall that a finite number of terms doesn’t
affect convergence.
 Therefore, Σ an is absolutely convergent.
THE RATIO TEST
If
Case ii—Proof
|an+1/an| → L > 1 or |an+1/an| → ∞
then the ratio |an+1/an| will eventually
be greater than 1.
 That is, there exists an integer N such that:
an 1
1
an
whenever n  N
Case ii—Proof
THE RATIO TEST
This means that |an+1| > |an| whenever
n ≥ N, and so
lim an  0
n 
 Therefore, Σan diverges by
the Test for Divergence.
Case iii—Proof
NOTE
Part iii of the Ratio Test says that,
if
lim an 1 / an  1
n 
the test gives no information.
Case iii—Proof
NOTE
For instance, for the convergent series Σ 1/n2,
we have:
an 1
an
1
2
2
n
(n  1)


1
(n  1) 2
2
n
1

1
2
 1
1  
 n
as n  
Case iii—Proof
NOTE
For the divergent series Σ 1/n,
we have:
an 1
an
1
n
n

1


1
n 1
n
1

1
1
1
n
as n  
NOTE
Case iii—Proof
Therefore, if lim an 1 / an  1,
n 
the series Σ an might converge
or it might diverge.
 In this case, the Ratio Test fails.
 We must use some other test.
Example 4
3
RATIO TEST
Test the series

n
(1) n

3
n 1
n
for absolute convergence.
 We use the Ratio Test with an = (–1)n n3 / 3n,
as follows.
Example 4
RATIO TEST
(1)
an 1

an
n 1
( n  1)
n 1
3
n 3
(1) n
n
3
3
(n  1) 3
 n 1  3
3
n
3
1  n 1 
 

3 n 
n
3
3
1 1
1
 1     1
3 n 
3
RATIO TEST
Example 4
Thus, by the Ratio Test, the given series
is absolutely convergent and, therefore,
convergent.
Example 5
RATIO TEST
Test the convergence of the series

n
n

n 1 n !
 Since the terms an = nn/n! are positive,
we don’t need the absolute value signs.
Example 5
RATIO TEST
n 1
an 1 (n  1)
n ! (n  1)(n  1) n !

 n 
 n
an
(n  1)! n
(n  1)n !
n
n
 n 1 


 n 
n
n
 1
 1    e
 n
as n  
 See Equation 6 in Section 3.6
 Since e > 1, the series is divergent by the Ratio Test.
NOTE
Although the Ratio Test works in Example 5,
an easier method is to use the Test for
Divergence.
n
n  n  n  n
an 

n
n! 1 2  3  n
n
 Since
it follows that an does not approach 0 as n → ∞.
 Thus, the series is divergent by the Test for Divergence.
ABSOLUTE CONVERGENCE
The following test is convenient to apply
when nth powers occur.
 Its proof is similar to the proof of the Ratio Test
and is left as Exercise 37.
Case i
THE ROOT TEST
If
lim n an  L  1
n 
then the series

 a is absolutely convergent
n 1
n
(and therefore convergent).
Case ii
THE ROOT TEST
If
lim n an  L  1 or lim n an  
n 
n

then the series
 a is divergent.
n 1
n
THE ROOT TEST
If
Case iii
lim n an  1
n 
the Root Test is inconclusive.
ROOT TEST
If lim n an  1, then part iii
n 
of the Root Test says that the test
gives no information.
 The series Σ an could converge or diverge.
ROOT TEST VS. RATIO TEST
If L = 1 in the Ratio Test, don’t try the Root
Test—because L will again be 1.
If L = 1 in the Root Test, don’t try the Ratio
Test—because it will fail too.
Example 6
ROOT TEST
Test the convergence of the series
 2n  3 



n 1  3n  2 

 2n  3 
an  

 3n  2 
n
n
3
2
2n  3
2
n
an 

 1
3n  2 3  2
3
n
 Thus, the series converges by the Root Test.
n
REARRANGEMENTS
The question of whether a given convergent
series is absolutely convergent or conditionally
convergent has a bearing on the question of
whether infinite sums behave like finite sums.
REARRANGEMENTS
If we rearrange the order of the terms
in a finite sum, then of course the value
of the sum remains unchanged.
 However, this is not always the case for
an infinite series.
REARRANGEMENT
By a rearrangement of an infinite series
Σ an, we mean a series obtained by simply
changing the order of the terms.
 For instance, a rearrangement of Σ an
could start as follows:
a1 + a2 + a5 + a3 + a4 + a15 + a6 + a7 + a20 + …
REARRANGEMENTS
It turns out that, if Σ an is an absolutely
convergent series with sum s, then
any rearrangement of Σ an has the same
sum s.
REARRANGEMENTS
However, any conditionally
convergent series can be rearranged
to give a different sum.
REARRANGEMENTS
Equation 6
To illustrate that fact, let’s consider
the alternating harmonic series
1  12  13  14  15  16  71  81  ...  ln 2
 See Exercise 36 in Section 11.5
REARRANGEMENTS
If we multiply this series by ½,
we get:
1
2
    ...  ln 2
1
4
1
6
1
8
1
2
Equation 7
REARRANGEMENTS
Inserting zeros between the terms of this
series, we have:
0   0   0   0   ...  ln 2
1
2
1
4
1
6
1
8
1
2
Equation 8
REARRANGEMENTS
Now, we add the series in Equations 6 and 7
using Theorem 8 in Section 11.2:
1       ...  ln 2
1
3
1
2
1
5
1
7
1
4
3
2
REARRANGEMENTS
Notice that the series in Equation 8
contains the same terms as in Equation 6,
but rearranged so that one negative term
occurs after each pair of positive terms.
REARRANGEMENTS
However, the sums of these series are
different.
 In fact, Riemann proved that, if Σ an is a conditionally
convergent series and r is any real number whatsoever,
then there is a rearrangement of Σ an that has a sum
equal to r.
 A proof of this fact is outlined in Exercise 40.