11.6 Absolute Convergence and the Ratio and Root tests

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


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


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


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


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.


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.


Then,

is the difference of two convergent series and is,

therefore, convergent.

( )n n n na a a a



Determine whether the series

is convergent or divergent.

Example 3

2 2 2 21

cos cos1 cos 2 cos3...

1 2 3n

n

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

2 2 2 21

cos cos1 cos 2 cos3...

1 2 3n

n

n


We can apply the Comparison Test to the series of

absolute values:

Example 3

2 21 1

coscos

n n

nn

n n


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


Thus, the given series (cos n)/n2 is absolutely

convergent and therefore, convergent by Theorem

3.

Example 3


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


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.

Documents

11.6 Absolute Convergence and the Ratio and Root tests