12INFINITE SEQUENCES AND SERIESINFINITE SEQUENCES AND SERIES

12.4The Comparison Tests

In this section, we will learn:

How to find the value of a series

by comparing it with a known series.

INFINITE SEQUENCES AND SERIES

COMPARISON TESTS

In the comparison tests, the idea

is to compare a given series with

one that is known to be convergent

or divergent.

Consider the series

This reminds us of the series .

The latter is a geometric series with a = ½ and r = ½ and is therefore convergent.

1

1

2 1nn

∞

= +∑

11/ 2n

n

∞

=∑

COMPARISON TESTS Series 1

As the series is similar to a convergent

series, we have the feeling that it too

must be convergent.

Indeed, it is.

COMPARISON TESTS

The inequality

shows that our given series has smaller terms

than those of the geometric series.

Hence, all its partial sums are also smaller than 1 (the sum of the geometric series).

1 1

2 1 2n n<

+

COMPARISON TESTS

Thus,

Its partial sums form a bounded increasing sequence, which is convergent.

It also follows that the sum of the series is less than the sum of the geometric series:

1

11

2 1nn

∞

=

<+∑

COMPARISON TESTS

Similar reasoning can be used to prove

the following test—which applies only to

series whose terms are positive.

COMPARISON TESTS

The first part says that, if we have a series

whose terms are smaller than those of

a known convergent series, then our series

is also convergent.

COMPARISON TESTS

The second part says that, if we start with

a series whose terms are larger than those

of a known divergent series, then it too is

divergent.

COMPARISON TESTS

Suppose that Σ an and Σ bn are series

with positive terms.

i. If Σ bn is convergent and an ≤ bn for all n, then Σ an is also convergent.

ii. If Σ bn is divergent and an ≥ bn for all n, then Σ an is also divergent.

THE COMPARISON TEST

Let

Since both series have positive terms, the sequences {sn} and {tn} are increasing (sn+1 = sn + an+1 ≥ sn).

Also, tn → t; so tn ≤ t for all n.

1 1 1

n n

n i n i ni i n

s a t b t b∞

= = =

= = =∑ ∑ ∑

THE COMPARISON TEST—PROOF Part i

Since ai ≤ bi, we have sn ≤ tn.

Hence, sn ≤ t for all n.

This means that {sn} is increasing and bounded above.

So, it converges by the Monotonic Sequence Theorem.

Thus, Σ an converges.

THE COMPARISON TEST—PROOF Part i

If Σ bn is divergent, then tn → ∞

(since {tn} is increasing).

However, ai ≥ bi; so sn ≥ tn.

Thus, sn → ∞; so Σ an diverges.

THE COMPARISON TEST—PROOF Part ii

It is important to keep in mind

the distinction between a sequence

and a series.

A sequence is a list of numbers.

A series is a sum.

SEQUENCE VS. SERIES

With every series Σ an, there are

associated two sequences:

1. The sequence {an} of terms

2. The sequence {sn} of partial sums

SEQUENCE VS. SERIES

In using the Comparison Test, we must,

of course, have some known series Σ bn

for the purpose of comparison.

COMPARISON TEST

Most of the time, we use one of

these:

A p-series [Σ 1/np converges if p > 1 and diverges if p ≤ 1]

A geometric series [Σ arn–1 converges if |r| < 1 and diverges if |r| ≥ 1]

COMPARISON TEST

Determine whether the given series

converges or diverges:

21

5

2 4 3n n n

∞

= + +∑

COMPARISON TEST Example 1

For large n, the dominant term in

the denominator is 2n2.

So, we compare the given series with the series Σ 5/(2n2).

COMPARISON TEST Example 1

Observe that

since the left side has a bigger denominator.

In the notation of the Comparison Test, an is the left side and bn is the right side.

2 2

5 5

2 4 3 2n n n<

+ +

COMPARISON TEST Example 1

We know that

is convergent because it’s a constant

times a p-series with p = 2 > 1.

2 21 1

5 5 1

2 2n nn n

∞ ∞

= =

=∑ ∑

COMPARISON TEST Example 1

Therefore,

is convergent by part i of the Comparison

Test.

COMPARISON TEST Example 1

21

5

2 4 3n n n

∞

= + +∑

Although the condition an ≤ bn or an ≥ bn

in the Comparison Test is given for all n,

we need verify only that it holds for n ≥ N,

where N is some fixed integer.

This is because the convergence of a series is not affected by a finite number of terms.

This is illustrated in the next example.

NOTE 1

Test the given series for convergence

or divergence:

1

ln

n

n

n

∞

=∑

COMPARISON TEST Example 2

This series was tested (using the Integral

Test) in Example 4 in Section 11.3

However, it is also possible to test it by comparing it with the harmonic series.

COMPARISON TEST Example 2

Observe that ln n > 1 for n ≥ 3.

So,

We know that Σ 1/n is divergent (p-series with p = 1).

Thus, the series is divergent by the Comparison Test.

ln 13

nn

n n> ≥

COMPARISON TEST Example 2

The terms of the series being tested must

be smaller than those of a convergent

series or larger than those of a divergent

series.

If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, the Comparison Test doesn’t apply.

NOTE 2

For instance, consider

The inequality is useless

as far as the Comparison Test is concerned.

This is because Σ bn = Σ (½)n is convergent and an > bn.

1

1

2 1nn

∞

= −∑NOTE 2

1 1

2 1 2n n>

−

Nonetheless, we have the feeling that

Σ1/(2n -1) ought to be convergent because

it is very similar to the convergent geometric

series Σ (½)n.

In such cases, the following test can be used.

NOTE 2

Suppose that Σ an and Σ bn are series

with positive terms.

If

where c is a finite number and c > 0,

either both series converge or both diverge.

lim n

nn

ac

b→ ∞=

LIMIT COMPARISON TEST

Let m and M be positive numbers such

that m < c < M.

Since an/bn is close to c for large n, there is an integer N such that

when

and so when

n

n

n n n

am M n N

b

mb a Mb n N

< < >

< < >

LIMIT COMPARISON TEST—PROOF

If Σ bn converges, so does Σ Mbn.

Thus, Σ an converges by part i of the Comparison Test.

LIMIT COMPARISON TEST—PROOF

If Σ bn diverges, so does Σ mbn.

Thus, Σ an diverges by part ii of the Comparison Test.

LIMIT COMPARISON TEST—PROOF

Test the given series for convergence

or divergence:

1

1

2 1nn

∞

= −∑

COMPARISON TESTS Example 3

We use the Limit Comparison Test

with:

1 1

2 1 2n nn na b= =

−

COMPARISON TESTS Example 3

We obtain:

COMPARISON TESTS Example 3

This limit exists and Σ 1/2n is

a convergent geometric series.

Thus, the given series converges by the Limit Comparison Test.

COMPARISON TESTS Example 3

Determine whether the given series

converges or diverges:

2

51

2 3

5n

n n

n

∞

=

+

+∑

COMPARISON TESTS Example 4

The dominant part of the numerator is 2n2.

The dominant part of the denominator is n5/2.

This suggests taking:

2 2

5/ 2 1/ 25

2 3 2 2

5n n

n n na b

n nn

+= = =

+

COMPARISON TESTS Example 4

2 1/ 2

5

5/ 2 3/ 2

5

5

2 3lim lim .

25

2 3lim

2 53

2 2 0lim 1

5 2 0 12 1

n

n nn

n

n

a n n n

b n

n n

n

n

n

→ ∞ → ∞

→ ∞

→ ∞

+=

++

=+

+ += = =

++

COMPARISON TESTS Example 4

We obtain:

Σ bn = 2 Σ 1/n1/2 is divergent

(p-series with p = ½ < 1).

Thus, the given series diverges by the Limit Comparison Test.

COMPARISON TESTS Example 4

Notice that, in testing many series,

we find a suitable comparison series Σ bn

by keeping only the highest powers in

the numerator and denominator.

COMPARISON TESTS

ESTIMATING SUMS

We have used the Comparison Test

to show that a series Σ an converges by

comparison with a series Σ bn.

It follows that we may be able to estimate the sum Σ an by comparing remainders.

As in Section 11.3, we consider

the remainder

Rn = s – sn = an+1 + an+2 + …

For the comparison series Σ bn, we consider the corresponding remainder

Tn = t - tn = bn+1 + bn+2 + …

ESTIMATING SUMS

As an ≤ bn for all n, we have Rn ≤ Tn.

If Σ bn is a p-series, we can estimate its remainder Tn as in Section 11.3

If Σ bn is a geometric series, then Tn is the sum of a geometric series and we can sum it exactly (Exercises 35 and 36).

In either case, we know that Rn is smaller than Tn

ESTIMATING SUMS

Use the sum of the first 100 terms to

approximate the sum of the series Σ 1/(n3+1).

Estimate the error involved in this

approximation.

ESTIMATING SUMS Example 5

Since

the given series is convergent by

the Comparison Test.

3 3

1 1

1n n<

+

ESTIMATING SUMS Example 5

The remainder Tn for the comparison

series Σ 1/n3 was estimated in Example 5

in Section 11.3 (using the Remainder

Estimate for the Integral Test).

We found that:

ESTIMATING SUMS Example 5

3 2

1 1

2n nT dx

x n

∞≤ =∫

Therefore, the remainder for

the given series satisfies:

Rn ≤ Tn ≤ 1/2n2

ESTIMATING SUMS Example 5

With n = 100, we have:

With a programmable calculator or a computer, we find that

with error less than 0.00005

100 2

10.00005

2(100)R ≤ =

100

3 31 1

1 10.6864538

1 1n nn n

∞

= =

≈ ≈+ +∑ ∑

ESTIMATING SUMS Example 5