11.4 The Comparison Tests

11.4 The Comparison Tests

The Comparison Tests

In the comparison tests the idea is to compare a given series with a series that is known to be convergent or divergent. For instance, the series

reminds us of the series , which is a geometric series with and and is therefore convergent.

Because the series (1) is so similar to a convergent series, we have the feeling that it too must be convergent. Indeed, it is.

The Comparison Tests

The inequality

shows that our given series (1) has smaller terms than those of the geometric series and therefore all its partial sums are also smaller than 1 (the sum of the geometric series).

This means that 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:

The Comparison Tests

Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive. 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.

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.

The Comparison Tests

In using the Comparison Test we must, of course, have some known series Σ b_n for the purpose of comparison.

Most of the time we use one of these series:

• A p-series [Σ 1/n^p converges if p > 1 and diverges if p ≤ 1]

• A geometric series [Σ ar^{n – 1} converges if | r| < 1 and diverges if | r | ≥ 1]

6

Example 1

Determine whether the series converges or diverges.

Solution:

For large n the dominant term in the denominator is 2n², so we compare the given series with the series Σ 5/(2n²).

Observe that

because the left side has a bigger denominator.

(In the notation of the Comparison Test, a_n is the left side and b is the right side.)

Example 1 – Solution

We know that

is convergent because it’s a constant times a p-series with p = 2 > 1.

Therefore

is convergent by part (i) of the Comparison Test.

cont’d

The Comparison Tests

Note 1:

Although the condition a_n ≤ b_n or a_n ≥ b_n 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, because the

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

The Comparison Tests

Note 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, then the

Comparison Test doesn’t apply.

The Comparison Tests

Consider, for instance, the series

The inequality

is useless as far as the Comparison Test is concerned because is convergent and a_n > b_n.

The Comparison Tests

Nonetheless, we have the feeling that Σ 1/(2ⁿ – 1) ought to be convergent because it is very similar to the convergent geometric series .

In such cases the following test can be used.

Example 3

Test the series for convergence or divergence.

Solution:

We use the Limit Comparison Test with

and obtain

Example 3 – Solution

Since this limit exists and Σ 1/2ⁿ is a convergent geometric series, the given series converges by the Limit Comparison Test.

cont’d

Estimating Sums

Estimating Sums

If we have used the Comparison Test to show that a series Σ a_n converges by comparison with a series Σ b_n, then we may be able to estimate the sum Σ a_n by comparing

remainders.

We consider the remainder

R_n = s – s_n = a_{n + 1} + a_{n + 2} + ^{. . .}

Estimating Sums

For the comparison series Σ b_n we consider the corresponding remainder

T_n = t – t_n = b_{n + 1} + b_{n + 2} + ^{. . .}

Since a_n ≤ b_n for all n, we have R_n ≤ T_n. If Σ b_n is a p-series, we can estimate its remainder T_n. If Σ b_n is a geometric

series, then T_n is the sum of a geometric series and we can sum it exactly.

In either case we know that R_n is smaller than T_n.

Example 5

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.

Solution:

Since

the given series is convergent by the Comparison Test.

Example 5 – Solution

The remainder T_n for the comparison series Σ 1/n³ was estimated earlier using the Remainder Estimate for the Integral Test.

There we found that

Therefore the remainder R_n for the given series satisfies

cont’d

Example 5 – Solution

With n = 100 we have

Using a programmable calculator or a computer, we find that

with error less than 0.00005.

cont’d