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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.
INFINITE SEQUENCES AND SERIES
ABSOLUTE CONVERGENCE
Given any series an, we can consider the
corresponding series
whose terms are the absolute values of the terms of
the original series.
1 2 31
...nn
a a a a
ABSOLUTE CONVERGENCE
A series an is called absolutely convergent if the
series of absolute values |an| is convergent.
Notice that, if an is a series with positive terms,
then |an| = an.
So, in this case, absolute convergence is the same as convergence.
Definition 1
ABSOLUTE CONVERGENCE
The series
is absolutely convergent because
is a convergent p-series (p = 2).
Example 1
1
2 2 2 21
( 1) 1 1 11 ...
2 3 4
n
n n
1
2 2 2 2 21 1
( 1) 1 1 1 11 ...
2 3 4
n
n nn n
ABSOLUTE CONVERGENCE
We know that the alternating harmonic series
is convergent.
See Example 1 in Section 11.5.
Example 2
1
1
( 1) 1 1 11 ...
2 3 4
n
n n
However, it is not absolutely convergent because
the corresponding series of absolute values is:
This is the harmonic series (p-series with p = 1) and is, therefore, divergent.
ABSOLUTE CONVERGENCE Example 2
1
1 1
( 1) 1 1 1 11 ...
2 3 4
n
n nn n
CONDITIONAL CONVERGENCE
A series an is called conditionally convergent if it
is convergent but not absolutely convergent.
Definition 2
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 an is absolutely convergent, then it is
convergent.
Theorem 3
Observe that the inequality
is true because |an| is either an or –an.
0 2n n na a a
ABSOLUTE CONVERGENCE Theorem 3 - Proof
If an is absolutely convergent, then |an| is
convergent.
So, 2|an| is convergent.
Thus, by the Comparison Test, (an + |an|) is convergent.
ABSOLUTE CONVERGENCE Theorem 3 - Proof
Then,
is the difference of two convergent series and is,
therefore, convergent.
( )n n n na a a a
ABSOLUTE CONVERGENCE Theorem 3 - Proof
ABSOLUTE CONVERGENCE
Determine whether the series
is convergent or divergent.
Example 3
2 2 2 21
cos cos1 cos 2 cos3...
1 2 3n
n
n
ABSOLUTE CONVERGENCE
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
2 2 2 21
cos cos1 cos 2 cos3...
1 2 3n
n
n
ABSOLUTE CONVERGENCE
We can apply the Comparison Test to the series of
absolute values:
Example 3
2 21 1
coscos
n n
nn
n n
ABSOLUTE CONVERGENCE
Since |cos n| 1 for all n, we have:
We know that 1/n2 is convergent (p-series with p = 2).
Hence, (cos n)/n2 is convergent by the Comparison Test.
Example 3
2 2
cos 1n
n n
ABSOLUTE CONVERGENCE
Thus, the given series (cos n)/n2 is absolutely
convergent and therefore, convergent by Theorem
3.
Example 3
ABSOLUTE CONVERGENCE
The following test is very useful in determining
whether a given series is absolutely convergent.
THE RATIO TEST
If
then the series is absolutely convergent
(and therefore convergent).
1lim 1n
nn
aL
a
1n
n
a
Case i
THE RATIO TEST
If
then the series is divergent.
1 1lim 1 or limn n
n nn n
a aL
a a
Case ii
1n
n
a
THE RATIO TEST
If
the Ratio Test is inconclusive.
That is, no conclusion can be drawn about the convergence or divergence of an.
1lim 1n
nn
a
a
Case iii
THE RATIO TEST
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
Since
the ratio |an+1/an| will eventually be less than r.
That is, there exists an integer N such that:
THE RATIO TEST
1lim and n
nn
aL L r
a
1 whenever n
n
ar n N
a
Case i - Proof
Equivalently,
|an+1| < |an|r whenever n N
THE RATIO TEST i-Proof (Inequality 4)
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 Case i - Proof
In general,
|aN+k| < |aN|rk for all k 1
THE RATIO TEST i-Proof (Inequality 5)
Now, the series
is convergent because it is a geometric series with
0 < r < 1.
THE RATIO TEST
2 3
1
...kN N N N
k
a r a r a r a r
Case i - Proof
Thus, inequality 5, together with the Comparison
Test, shows that the series
is also convergent.
THE RATIO TEST
1 2 31 1
...n N k N N Nn N k
a a a a a
Case i - Proof
THE RATIO TEST
It follows that the series is convergent.
Recall that a finite number of terms does not affect
convergence.
Therefore, an is absolutely convergent.
1n
n
a
Case i - Proof
THE RATIO TEST
If |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:
1 1 whenever n
n
an N
a
Case ii - Proof
THE RATIO TEST
This means that |an+1| > |an| whenever n N, and so
Therefore, an diverges by the Test for Divergence.lim 0nn
a
Case ii - Proof
NOTE
Part iii of the Ratio Test says that,
if
the test gives no information.
1lim / 1n nn
a a
Case iii - Proof
NOTE
For instance, for the convergent series 1/n2, we
have:22
12
2
2
1( 1)
1 ( 1)
11 as
11
n
n
a nna n
n
n
n
Case iii - Proof
NOTE
For the divergent series 1/n, we have:
1
11
1 1
11 as
11
n
n
a nna n
n
n
n
Case iii - Proof
NOTE
Therefore, if , the series an
might converge or it might diverge.
In this case, the Ratio Test fails.
We must use some other test.
1lim / 1n nn
a a
Case iii - Proof
RATIO TEST
Test the series
for absolute convergence.
We use the Ratio Test with an = (–1)n n3 / 3n, as follows.
Example 4
3
1
( 1)3
nn
n
n
RATIO TEST Example 4
1 3
3311
3 1 3
3
( 1) ( 1)( 1) 3 1 13
( 1) 3 33
1 1 11 1
3 3
n
nnn
n nn
n
na n n
na n n
n
RATIO TEST Example 4
Therefore, by the Ratio Test, the given series is
absolutely convergent and, therefore, convergent.
RATIO TEST
Test the convergence of the series
Since the terms an = nn/n! are positive, we do not need the absolute value signs.
Example 5
1 !
n
n
n
n
RATIO TEST
See Equation 6 in Section 3.6 Since e > 1, the series is divergent by the Ratio Test.
Example 5
11 ( 1) ! ( 1)( 1) !
( 1)! ( 1) !
1
11 as
n nn
n nn
n
n
a n n n n n
a n n n n n
n
n
e nn
NOTE
Although the Ratio Test works in Example 5, an
easier method is to use the Test for Divergence.
Since
it follows that an does not approach 0 as n → .
Thus, the series is divergent by the Test for Divergence.
! 1 2 3
n
n
n n n n na n
n n
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.
THE ROOT TEST
If
then the series is absolutely convergent
(and therefore convergent).
lim 1nn
na L
1n
n
a
Case i
THE ROOT TEST
If
then the series is divergent.
lim 1 or limn nn nn n
a L a
Case ii
1n
n
a
THE ROOT TEST
If
the Root Test is inconclusive.
lim 1nn
na
Case iii
ROOT TEST
If , then part iii of the Root Test
says that the test gives no information.
The series an could converge or diverge.
lim 1nnn
a
ROOT TEST VS. RATIO TEST
If L = 1 in the Ratio Test, do not try the Root Test,
because L will again be 1.
If L = 1 in the Root Test, do not try the Ratio Test,
because it will fail too.
ROOT TEST
Test the convergence of the series
Thus, the series converges by the Root Test.
Example 6
1
2 3
3 2
n
n
n
n
2 3
3 2
322 3 2
123 2 33
n
n
nn
na
n
n nan
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 an, we
mean a series obtained by simply changing the
order of the terms.
For instance, a rearrangement of an could start as follows:
a1 + a2 + a5 + a3 + a4 + a15 + a6 + a7 + a20 + …
REARRANGEMENTS
It turns out that, if an is an absolutely convergent
series with sum s, then any rearrangement of an
has the same sum s.
REARRANGEMENTS
However, any conditionally convergent series can
be rearranged to give a different sum.
REARRANGEMENTS
To illustrate that fact, let us consider the
alternating harmonic series
See Exercise 36 in Section 11.5
1 1 1 1 1 1 12 3 4 5 6 7 81 ... ln 2
Equation 6
REARRANGEMENTS
If we multiply this series by ½, we get:
1 1 1 1 12 4 6 8 2... ln 2
REARRANGEMENTS
Inserting zeros between the terms of this series, we
have:
Equation 7
1 1 1 1 12 4 6 8 20 0 0 0 ... ln 2
REARRANGEMENTS
Now, we add the series in Equations 6 and 7 using
Theorem 8 in Section 11.2:
Equation 8
31 1 1 1 13 2 5 7 4 21 ... ln 2
1 1 1 1 1 1 12 3 4 5 6 7 8
1 1 1 1 12 4 6 8 2
1 ... ln 2
0 0 0 0 ... ln 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 an is a conditionally convergent series and r is any real number whatsoever, then there is a rearrangement of an that has a sum equal to r.
A proof of this fact is outlined in Exercise 40.