Infinite Sequences and Series

The Integral and Comparison Tests; Estimating Sums

8.3

The Integral and Comparison Tests; Estimating Sums

In general, it is difficult to find the exact sum of a series. We

were able to accomplish this for geometric series and the

series  1/[n(n + 1)] because in each of those cases we

could find a simple formula for the nth partial sum s

But usually it isn’t easy to compute lim

n→

Therefore in this section we develop tests that enable us to

determine whether a series is convergent or divergent

without explicitly finding its sum.

Testing with an Integral

Let’s investigate the series whose terms are the reciprocals

of the squares of the positive integers:

There’s no simple formula for

the sum s

of the first n terms,

but the computer-generated

table of values given to the right

suggests that the partial sums

are approaching a number

near 1.64 as n → and so it

looks as if the series is convergent.

Testing with an Integral

We can confirm this impression with a geometric argument.

Figure 1 shows the curve y = 1/x

and rectangles that lie

below the curve.

The base of each rectangle is an interval of length 1; the

height is equal to the value of the function y = 1/x

at the

right endpoint of the interval.

Figure 1

Testing with an Integral

So the sum of the areas of the rectangles is

If we exclude the first rectangle, the total area of the

remaining rectangles is smaller than the area under the

curve y = 1/x

for x  1, which is the value of the

integral

The improper integral is convergent and has value 1. So the

picture shows that all the partial sums are less than

Testing with an Integral

Thus the partial sums are bounded and the series

converges. The sum of the series (the limit of the partial

sums) is also less than 2:

Now let’s look at the series

Testing with an Integral

The table of values of s

suggests that the partial sums

aren’t approaching a finite number, so we suspect that the

given series may be divergent.

Testing with an Integral

Again we use a picture for confirmation. Figure 2 shows the

curve but this time we use rectangles whose tops

lie above the curve.

The base of each rectangle is an interval of length 1. The

height is equal to the value of the function at the

left endpoint of the interval.

Figure 2

Testing with an Integral

So the sum of the areas of all the rectangles is

This total area is greater than the area under the curve

for x  1, which is equal to the integral

But we know that this improper integral is divergent. So the

sum of the series must be infinite, that is, the series is

divergent.

Testing with an Integral

The same sort of geometric reasoning that we used for

these two series can be used to prove the following test.

Example 1 – Using the Integral Test

Determine whether the series converges or

diverges.

Solution:

The function f(x) = (ln x)/x is positive and continuous for

x > 1 because the logarithm function is continuous.

But it is not obvious whether or not f is decreasing, so we

compute its derivative:

Example 1 – Solution

Thus f'(x) < 0 when ln x > 1, that is, x > e. It follows that f is

decreasing when x > e and so we can apply the Integral

Test:

Since this improper integral is divergent, the series  (ln n)/n

is also divergent by the Integral Test.

Testing with an Integral

The series is called the p-series.

For instance, the series

is convergent because it is a p-series with p = 3 > 1. But the

series

is divergent because it is a p-series with

Testing by Comparing

The series

reminds us of the series which is a geometric

series with a = and r = and is therefore convergent.

Because the series (2) is so similar to a convergent series,

we have the feeling that it too must be convergent. Indeed, it

is. The inequality

shows that our given series (2) 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).

Testing by Comparing

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:

Similar reasoning can be used to prove the following test,

which applies only to series whose terms are positive.

Testing by Comparing

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.

Testing by Comparing

Standard Series for Use with the Comparison Test

In using the Comparison Test we must, of course, have

some known series  b

for the purpose of comparison.

Most of the time we use one of these series:

▪ A p-series [ 1/n

converges if p > 1 and diverges if p  1;

see (1)]

▪ A geometric series [ ar

n–1

converges if |r| < 1 and

diverges if |r|  1]

Example 3 – Using the Comparison Test

Determine whether the series converges

or diverges.

Solution:

For large n the dominant term in the denominator is 2n

, so

we compare the given series with the series  5/(2n

Observe that

because the left side has a bigger denominator. (In the

notation of the Comparison Test, a

is the left side and b

the right side.)

Example 3 – Solution

We know that

is convergent because it’s a constant times a p-series with

p = 2 > 1.

Therefore

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

cont’d

Testing by Comparing

Consider the series

The inequality

is useless as far as the Comparison Test is concerned

because  b

=  is convergent and a

> b

Nonetheless, we have the feeling that  1/(2

– 1) ought to

be convergent because it is very similar to the convergent

geometric series  .

Testing by Comparing

In such cases the following test can be used.

Estimating the Sum of a Series

Suppose we have been able to use the Integral Test to show

that a series  a

is convergent and we now want to find an

approximation to the sum s of the series.

Of course, any partial sum s

is an approximation to s

because lim

n→

= s. But how good is such an

approximation? To find out, we need to estimate the size of

the remainder

= s – s

= a

n+1

+ a

n+2

+ a

n+3

The remainder R

is the error made when s

, the sum of the

first n terms, is used as an approximation to the total sum.

Estimating the Sum of a Series

We use the same notation and ideas as in the Integral Test,

assuming that f is decreasing on [n, ). Comparing the

areas of the rectangles with the area under y = f(x) for x > n

in Figure 3, we see that

Similarly, we see from Figure 4 that

Figure 3

Figure 4

Estimating the Sum of a Series

So we have proved the following error estimate.

Example 6 – Estimating the Sum of a Series

(a) Approximate the sum of the series  1/n

by using the

sum of the first 10 terms. Estimate the error involved in

this approximation.

(b) How many terms are required to ensure that the sum is

accurate to within 0.0005?

Solution:

In both parts (a) and (b) we need to know With

f(x) = 1/x

, which satisfies the conditions of the Integral Test,

we have

Example 6 – Solution

(a) Approximating the sum of the series by the 10th partial

sum, we have

According to the remainder estimate in (3), we have

So the size of the error is at most 0.005.

cont’d

Example 6 – Solution

(b) Accuracy to within 0.0005 means that we have to find a

value of n such that R

 0.0005. Since

we want

Solving this inequality, we get

We need 32 terms to ensure accuracy to within 0.0005.

cont’d

Estimating the Sum of a Series

If we add s

to each side of the inequalities in (3), we get

because s

+ R

= s. The inequalities in (4) give a lower

bound and an upper bound for s.

They provide a more accurate approximation to the sum of

the series than the partial sum s

does.