12 INFINITE SEQUENCES AND SERIES. 12.6 Absolute Convergence and the Ratio and Root tests In this...
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Transcript of 12 INFINITE SEQUENCES AND SERIES. 12.6 Absolute Convergence and the Ratio and Root tests In this...
12.6Absolute 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 Σ 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 Σ an is called absolutely
convergent if the series of absolute
values Σ |an| is convergent.
Definition 1
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
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 Σ 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.
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 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,
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
Fig. 12.5.2, p. 747
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 Σ 1/n2 is convergent (p-series with p = 2).
Hence, Σ (cos n)/n2 is convergent by the Comparison Test.
Example 3
2 2
cos 1n
n n
ABSOLUTE CONVERGENCE
Thus, the given series Σ (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
the Ratio Test is inconclusive.
That is, no conclusion can be drawn about the convergence or divergence of Σ 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
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
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, the 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 doesn’t
affect convergence.
Therefore, Σ 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, Σan diverges by the Test for Divergence.
lim 0nna
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 Σ 1/n2,
we have:
221
2
2
2
1( 1)
1 ( 1)
11 as
11
n
n
a nna n
n
n
n
Case iii—Proof
NOTE
Therefore, if ,
the series Σ 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 41 3
311
3 1 3
3
3
( 1) ( 1)( 1) 33
( 1) 33
1 1
3
1 1 11 1
3 3
n
nnn
n nn
n
na n
na n
n
n
n
RATIO TEST Example 4
Thus, 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 don’t 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 51
1 ( 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
ROOT TEST
If , then part iii
of the Root Test says that the test
gives no information.
The series Σ an could converge or diverge.
lim 1nn
na
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.
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
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
To illustrate that fact, let’s 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
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
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.