## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 14: Infinite Series – Section 14.6: The Alternating-Series And Absolute-Convergence Tests

14.6
The Alternating-Series And Absolute-Convergence
Tests

 1. Alternating Series

Alternating Series

A series may have finitely many negative terms, and thus it must have infinitely many positive-or-0-valued terms; they're
ultimately positive series. A series may have infinitely many negative terms, and thus it may have either no or finitely or
infinitely many positive-or-0-valued terms. All the tests discussed in previous sections apply only to ultimately positive series. If
a series has infinitely many negative terms and no or finitely many positive terms, the tests can be applied to the series of the
opposite or negative of its terms, which is an (ultimately) positive series, and which converges iff the original series converges.

If a series has infinitely many negative terms and infinitely many positive terms, no previous test can be applied to it directly. In
this section we present two tests: the alternating-series test, which applies to alternating series, and the absolute-convergence
test, which applies to any kind of series.

The series:

Definition 1.1 – Alternating Series

 A series is said to be an alternating series if its terms alternate in sign.

The Alternating-Series Test (AST)

The harmonic series:

The second one is just the negative of the first one. One converges iff the other converges. Here we deal with the first one.
Does it diverge to infinity too, or does it diverge to negative infinity, or does it simply diverge, or does it converge? Refer to Fig.
1.1. Let an and sn be the nth term and nth partial sum respectively. Note that an = 1/n > 0 if n is odd and an = –1/n < 0 if n is

 Fig. 1.1   Partial Sums Of Alternating Harmonic Series.

even. We have:

etc. If n is odd, then an+1 = (–1)(n+1)+1/n = –1/n < 0, and sn jumps downward to sn+1 = sn + an+1 by an amount equal to the size
of the next term, namely |an+1|. If n is even, then an+1 = 1/n > 0, and sn jumps upward to sn+1 by an amount equal to the next
term, namely an+1. Since an approaches 0 as n approaches infinity, the size of the oscillation of sn as a function of n is
decreasing toward 0. In fact for any positive integer n we get:

s2 < s4 < s6 < s8 < s10 < ... < s2n–2 < s2n < s2n–1 < s2n–3 < ... < s9 < s7 < s5 < s3 < s1.

Each even-numbered partial sum is less than every odd-numbered partial sum. The sequence of even-numbered partial sums
is increasing. The sequence of odd-numbered partial sums is decreasing. As the size of the oscillation back and forth between
the two sequences approaches 0, the gap between them approaches 0 as n approaches infinity, and thus they converge to the
same number. Consequently the sequence of partial sums and hence the alternating harmonic series converge to that number.

The following test asserts that a series converges provided it's an alternating series and satisfies two more conditions. It's
called the alternating-series test, abbreviated as “AST”.

Theorem 1.1 – The Alternating-Series Test (AST)

 then it converges.

Proof
Suppose a1 > 0. Then an > 0 if n is odd and an < 0 if n is even. Let sn be the nth partial sum. For any n > 0 we have:

The proof for the case where a1 < 0 is similar.
EOP

The Ultimately Version

Theorem 1.1 is also valid for series that's ultimately alternating and/or where the sequence of the absolute values of its terms
is ultimately decreasing.

The Ultimately Version Of The Alternating-Series Test

 then it converges.

Proof

EOP

Remarks 1.1

1. For the decreasing condition, we have to consider the absolute values, since the terms themselves are neither decreasing
nor increasing.

2. A sequence that's decreasing in absolute values doesn't necessarily approach 0. For example, the alternating harmonic
sequence has the absolute values of its terms decreasing and its terms approaching 0, while:

has its terms approaching 0 but their absolute values neither decreasing nor increasing.

Example 1.1

Show that the alternating harmonic series converges.

Solution
The sequence of the absolute values of its terms is decreasing and the sequence of its terms approaches 0. So it converges.
EOS

Example 1.2

Discuss the behavior (convergence or divergence) of the following series:

Solution

a. The series is alternating. The absolute values of its terms are decreasing. Its terms approach 0. So, by the AST, it
converges.

EOS

Example 1.3

Determine the convergence or divergence of this series:

Solution
The series is alternating. Let an = (–1)n+1 ( ln n)/n. We have:

converges, by the AST.
EOS

 2. Absolute Convergence

Absolute Convergence

Given a series, the series of the absolute values of its terms is a positive series, to which the SCT, LCT, RooT, RaT, and IT can
be applied. This fact motivates the investigation of absolute convergence.

Consider the series:

Definition 2.1 – Absolute Convergence

 A series is said to converge absolutely, or to be absolutely convergent, if the series of the absolute values of its terms converges:

The Absolute-Convergence Test (ACT)

For series that have infinitely many negative terms and infinitely many positive terms and that aren't alternating, the tests
discussed in previous sections and the AST can't be applied. Suppose a series converges absolutely. The sum of its terms is
less than or equal to the sum of the absolute values of the terms, which is a (finite) number because the series converges
absolutely. So the sum of the terms can't be infinity or negative infinity. Can the sum of the terms oscillate between two values
without the oscillation approaching 0 (eg, the sum oscillates between 100 and 102 as the terms oscillate between –2 and 2)?
No, because if it did, then the absolute values of the terms wouldn't approach 0 and the series wouldn't converge absolutely.
Thus the sum of the terms must be a (finite) number too. That is, the series converges too.

For a series that has infinitely many negative terms and infinitely many positive terms and that's not alternating, just like any
series, if it converges absolutely then it converges. In general, the cancellation that occurs because some terms are positive
and others are negative makes it easier for a series to converge than if all the terms are of one sign. The implication of the
convergence by the absolute convergence can be used to test the convergence of such series. Testing the convergence of a
series by examining its absolute convergence is known as the absolute-convergence test, or “ACT” for short.

Theorem 2.1 – The Absolute-Convergence Test (ACT)

 If a series converges absolutely then it converges.

Note On The Proof

original series = first defined series + second defined series.

The first defined series contains the positive terms and the second one contains the negative terms of the original series. For
the first one, we replace every negative term in the original series by 0, and for the second one, we replace every positive term
in the original series by 0. These replacements are made in order to have the original series as the sum of the two defined
series.

Proof

EOP

Remarks 2.1

1. Again the ACT can be used to test the convergence of a series that has infinitely many negative terms and infinitely many
positive terms and that's not alternating.

4. Every convergent positive series is also absolutely convergent, for it's identical to its corresponding series of absolute values.

Testing For Absolute Convergence

Example 2.1

Establish the behavior (convergence or divergence) of the series:

with the pattern that two positive terms are followed by one negative term, starting with two positive terms.

Note

That series has infinitely many negative terms and infinitely many positive terms. So the SCT, LCT, RooT, RaT, IT, and p-series
property can't be applied to it or to its opposite (negative). It's not alternating. Thus the AST can't be applied either. The
remaining hope is the ACT.

Solution
The series of the absolute values of the terms of the given series is:

which is a p-series with p = 3 > 1 and thus converges. Since the given series converges absolutely, it converges, by the ACT.
EOS

Conditional Convergence

Definition 2.2 – Conditional Convergence

 A series is said to converge conditionally, or to be conditionally convergent, if it converges but doesn't converge absolutely.

Remarks 2.2

1. As an example, the alternating harmonic series converges conditionally.

2. Conditional convergence and absolute convergence are mutually exclusive properties. If a series is conditionally convergent,
then it's not absolutely convergent, and if a series is absolutely convergent, then it's not conditionally convergent.

3. If a series doesn't converge absolutely, there remains the possibility that it converges conditionally. So again, we can't
conclude that a non-absolutely-convergent series is non-convergent. We have to try other tests to determine its convergence
or divergence.

4. If a series converges, then it converges either also absolutely or just conditionally.

5. If a series satisfies the AST then it converges. It doesn't necessarily converge conditionally. It converges conditionally only if
it doesn't converge absolutely.

Example 2.2

Solution
The corresponding series of absolute values is the harmonic series, which diverges. So the given series can't be absolutely
convergent. Clearly, by the AST, the given series converges. Thus the given series converges conditionally.
EOS

Example 2.3

Classify the series:

as absolutely convergent, conditionally convergent, or divergent.

Solution
Let an = (–2)n/n!. We then have:

EOS

As the two examples above show, when classifying a series as absolutely convergent, conditionally convergent, or divergent,
we should first check for absolute convergence. If it's absolutely convergent, we're done. If it's not, we check for convergence.
It's a waste of time to check for convergence first, because if it's convergent, we still have to check for absolute convergence.

Example 2.4

Classify the series:

as absolutely convergent, conditionally convergent, or divergent.

Solution

EOS

Example 2.5

Classify this series:

as absolutely convergent, conditionally convergent, or divergent.

Solution

EOS

 3. Series Whose Terms Involve Functions Of A Real Variable

Consider the series:

are members of the divergent family.

Example 3.1

For what values of x does the series:

converge absolutely? Converge conditionally? Diverge?

Solution
Let an = (x – 3)n/4nn. We have:

converges absolutely for –1 < x < 7, converges conditionally for x = –1, and diverges everywhere else.
EOS

Example 3.2

Determine the values of x for which this series:

converges absolutely, converges conditionally, or diverges.

Solution
Let an = (n + 3)2 (x/(x + 2))n. Then:

EOS

We can square both sides of |x| < |x + 2| and retain the direction of the inequality because absolute values are greater than
or equal to 0.

 4. Kinds Of Series And Tests

The tests that have been discussed in previous sections and in this section, in order of appearance, are the SCT, LCT, RooT,
RaT, IT, AST, and ACT. The p-series property has also been discussed.

1. For an ultimately positive series (finite number of negative terms), all the tests except the AST can be applied.

2. For an ultimately negative series (finite number of positive terms), all the tests except the AST can be applied to its negative,
which is an ultimately positive series.

3. For a series that's neither ultimately positive nor ultimately negative (infinitely many negative terms and infinitely many
positive terms) nor alternating, the ACT can be applied.

4. For an alternating series, the AST and ACT can be applied.

5. The ACT can be applied to any kind of series.

6. For a p-series, of course the p-series property can be applied.

# Problems & Solutions

1. Mark each of the following as T (true) or F (false).

Solution

2.
a. The following statement is false. Give a counter-example.

Solution

a. The alternating harmonic series:

3. Consider the series:

a. Does the AST apply directly to it?
b. Classify it as absolutely convergent, conditionally convergent, or divergent.

Solution

which obviously, by the AST, converges. Consequently the given series is conditionally convergent.

4. Classify the series:

as absolutely convergent, conditionally convergent, or divergent.

Solution

is convergent, by the AST, and consequently is conditionally convergent.

5. For what values of x does the series:

converge absolutely? Converge conditionally? Diverge?

Solution

Let an = xn/(2n ln n). Then:

So the given series converges absolutely for –2 < x < 2, converges conditionally for x = –2, and diverges everywhere else.

6. See the Proof Of Theorem 2.1 for this problem. Let:

Solution