INFINITE SEQUENCES AND SERIES

11INFINITE SEQUENCES AND SERIESINFINITE SEQUENCES AND SERIES

11.6Absolute 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.

ABSOLUTE CONVERGENCE

If a series Σ 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 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

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

then the series is divergent.

1 1lim 1 or limn n

n nn n

a aL

a a

1n

n

a

Case ii

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

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, 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

For the divergent series Σ 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 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

THE ROOT TEST

If

then the series is divergent.

lim 1 or limn nn nn na 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 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

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

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.