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APEX Calculus

Section 9.3 Integral and Comparison Tests

Knowing whether or not a series converges is very important, especially when we discuss Power Series in Section 9.6. Theorems 9.2.7 and 9.2.13 give criteria for when Geometric and p-series converge, and Theorem 9.2.23 gives a quick test to determine if a series diverges. There are many important series whose convergence cannot be determined by these theorems, though, so we introduce a set of tests that allow us to handle a broad range of series. We start with the Integral Test.

Subsection 9.3.1 Integral Test

We stated in Section 9.1 that a sequence {an} is a function a(n) whose domain is N, the set of natural numbers. If we can extend a(n) to R, the real numbers, and it is both positive and decreasing on [1,), then the convergence of n=1an is the same as 1a(x)dx.
Figure 9.3.2. Video presentation of Theorem 9.3.1
We can demonstrate the truth of the Integral Test with two simple graphs. In Figure 9.3.3.(a), the height of each rectangle is a(n)=an for n=1,2,, and clearly the rectangles enclose more area than the area under y=a(x). Therefore we can conclude that
(9.3.1)1a(x)dx<n=1an.
First of two graphs illustrating why the integral test works.
The graph of a positive, decreasing function y=a(x) is shown. Also shown are the first four rectangles in a Riemann sum with partition points at each integer, and using left endpoints. Since a(x) is decreasing, this is an overestimate, which is illustrated by the fact that the rectangles all reach above the graph.
(a)
Second of two graphs illustrating why the integral test works.
The graph of a positive, decreasing function y=a(x) is shown. Also shown are the first four rectangles in a Riemann sum with partition points at each integer, and using right endpoints. Since a(x) is decreasing, this is an underestimate, which is illustrated by the fact that the rectangles all lie below the graph.
(b)
Figure 9.3.3. Illustrating the truth of the Integral Test
In Figure 9.3.3.(b), we draw rectangles under y=a(x) with the Right-Hand rule, starting with n=2. This time, the area of the rectangles is less than the area under y=a(x), so n=2an<1a(x)dx. Note how this summation starts with n=2; adding a1 to both sides lets us rewrite the summation starting with n=1:
(9.3.2)n=1an<a1+1a(x)dx.
Combining Equations (9.3.1) and (9.3.2), we have
(9.3.3)n=1an<a1+1a(x)dx<a1+n=1an.
From Equation (9.3.3) we can make the following two statements:
  1. If n=1an diverges, so does 1a(x)dx (because n=1an<a1+1a(x)dx)
  2. If n=1an converges, so does 1a(x)dx (because 1a(x)dx<n=1an.)
Therefore the series and integral either both converge or both diverge. Theorem 9.2.24 allows us to extend this theorem to series where a(n) is positive and decreasing on [b,) for some b>1. A formal proof of the Integral Test is shown below.

Proof of the Integral Test.

Let a(x)=ax be a postive, continuous, decreasing function on [1,). We will consider how the partial sums of n=1an compare to the integral 0a(x)dx. We first consider the case where 1a(x)dx diverges.
  1. Suppose that 1a(x)dx diverges. Using Figure 9.3.3.(a), we can say that Sn=i=1nai>1n+1a(x)dx. If we let n in this inequality, we know that 1n+1a(x)dx will get arbitrarily large as n (since a(x)>0 and 1a(x)dx diverges). Therefore we conclude that Sn=i=1nai will also get arbitrarily large as n, and thus n=1an diverges.
  2. Now suppose that 1a(x)dx converges to M, where M is some positive, finite number. Using Figure 9.3.3.(b), we can say that 0<Sn=i=1nai<1a(x)dx=M. Therefore our sequence of partial sums, Sn is bounded. Furthermore, Sn is a monotonically increasing sequence since all of the terms an are positive. Since Sn is both bounded and monotonic, Sn converges by Convergent Sequences are Bounded and by Definition 9.2.2, the series n=1an converges as well.

Example 9.3.4. Using the Integral Test.

Determine the convergence of n=1ln(n)n2. (The terms of the sequence {an}={ln(n)/n2} and the nth partial sums are given in Figure 9.3.5.)
Solution 1.
Figure 9.3.5 implies that a(n)=(ln(n))/n2 is positive and decreasing on [2,). We can determine this analytically, too. We know a(n) is positive as both ln(n) and n2 are positive on [2,). Treating a(n) as a continuous function of n defined on [1,), consider a(n)=(12ln(n))/n3, which is negative for n2. Since a(n) is negative, a(n) is decreasing for n2. We can still use the integral test since a finite number of terms will not affect convergence of the series.
Scatter plots of the sequence used in this example, and the corresponding sequence of partial sums.
A scatter plot of the sequence an=ln(n)/n2 is shown. Aside from an initial point (1,0), this is a decreasing sequence, descending from the point (2,ln(2)/4) toward the n axis.
The scatter plot for the sequence Sn of partial sums is also shown; this is an increasing sequence, which appears to be bounded above by some value less than 1.
Figure 9.3.5. Plotting the sequence and series in Example 9.3.4
Applying the Integral Test, we test the convergence of 1ln(x)x2dx. Integrating this improper integral requires the use of Integration by Parts, with u=ln(x) and dv=1/x2dx.
1ln(x)x2dx=limb1bln(x)x2dx=limb1xln(x)|1b+1b1x2dx=limb1xln(x)1x|1b=limb11bln(b)b. Apply L'Hospital's Rule: =1.
Since 1ln(x)x2dx converges, so does n=1ln(n)n2.
Solution 2. Video solution
Theorem 9.2.13 was given without justification, stating that the general p-series n=11(an+b)p converges if, and only if, p>1. In the following example, we prove this to be true by applying the Integral Test.

Example 9.3.6. Using the Integral Test to establish Theorem 9.2.13.

Let a,b be real numbers such that a0 and an+b>0 for all n1. Use the Integral Test to prove that n=11(an+b)p converges if, and only if, p>1.
Solution 1.
Consider the integral 11(ax+b)pdx; assuming p1 and a0,
11(ax+b)pdx=limc1c1(ax+b)pdx=limc1a(1p)(ax+b)1p|1c=limc1a(1p)((ac+b)1p(a+b)1p).
This limit converges if, and only if, p>1 so that 1p<0. It is easy to show that the integral also diverges in the case of p=1. (This result is similar to the work preceding Key Idea 6.8.17.)
Therefore n=11(an+b)p converges if, and only if, p>1.
Solution 2. Video solution
We consider two more convergence tests in this section, both comparison tests. That is, we determine the convergence of one series by comparing it to another series with known convergence.

Subsection 9.3.2 Direct Comparison Test

Proof.

Let 0<anbn for all nN1. Note that both partial sums for both series are positive and increasing since the terms of both sequences are positive.
  1. Suppose that n=1bn converges, so n=1bn=S, where S is a finite, positive number. (S must be positive since bn>0.)
    Comparing the partial sums, we must have i=Nnaii=Nnbi since anbn for all nN. Furthermore since n=1bn converges to S, our partial sums for an are bounded (note that the partial sums started at i=N, but a finite number of terms will not affect the boundedness of the partial sums).
    0<i=Nnaii=Nnbi<S.
    Since the sequence of partial sums, sn=i=1nai is both monotonically increasing and bounded, we can say that sn converges (by Convergent Sequences are Bounded), and therefore so does n=1an.
  2. Suppose that n=1an diverges, so i=1nan=. (We can say that the series diverges to since the terms of the series are always positive). Comparing the partial sums, we have
    i=Nnaii=Nnbi
    Then applying limits, we get
    limni=Nnailimni=Nnbi.
    Since the limit on the left side diverges to , we can say that limni=Nnbi also diverges to .
Figure 9.3.8. Video presentation of Theorem 9.3.7

Example 9.3.9. Applying the Direct Comparison Test.

Determine the convergence of n=113n+n2.
Solution 1.
This series is neither a geometric or p-series, but seems related. We predict it will converge, so we look for a series with larger terms that converges. (Note too that the Integral Test seems difficult to apply here.)
Since 3n<3n+n2, 13n>13n+n2 for all n1. The series n=113n is a convergent geometric series; by Theorem 9.3.7, n=113n+n2 converges.
Solution 2. Video solution

Example 9.3.10. Applying the Direct Comparison Test.

Determine the convergence of n=11nln(n).
Solution 1.
We know the Harmonic Series n=11n diverges, and it seems that the given series is closely related to it, hence we predict it will diverge.
Since nnln(n) for all n1, 1n1nln(n) for all n1.
The Harmonic Series diverges, so we conclude that n=11nln(n) diverges as well.
Solution 2. Video solution
The concept of direct comparison is powerful and often relatively easy to apply. Practice helps one develop the necessary intuition to quickly pick a proper series with which to compare. However, it is easy to construct a series for which it is difficult to apply the Direct Comparison Test.
Consider n=11n+ln(n). It is very similar to the divergent series given in Example 9.3.10. We suspect that it also diverges, as 1n1n+ln(n) for large n. However, the inequality that we naturally want to use “goes the wrong way”: since nn+ln(n) for all n1, 1n1n+ln(n) for all n1. The given series has terms less than the terms of a divergent series, and we cannot conclude anything from this.
Fortunately, we can apply another test to the given series to determine its convergence.
Figure 9.3.11. Motivating Theorem 9.3.12

Subsection 9.3.3 Limit Comparison Test

Theorem 9.3.12 is most useful when the convergence of the series from {bn} is known and we are trying to determine the convergence of the series from {an}.
Figure 9.3.13. Video presentation of Theorem 9.3.12
We use the Limit Comparison Test in the next example to examine the series n=11n+ln(n) which motivated this new test.

Example 9.3.14. Applying the Limit Comparison Test.

Determine the convergence of n=11n+ln(n) using the Limit Comparison Test.
Solution 1.
We compare the terms of n=11n+ln(n) to the terms of the Harmonic Sequence n=11n:
limn1/(n+ln(n))1/n=limnnn+ln(n)=1 (after applying L'Hospital's Rule) .
Since the Harmonic Series diverges, we conclude that n=11n+ln(n) diverges as well.
Solution 2. Video solution

Example 9.3.15. Applying the Limit Comparison Test.

Determine the convergence of n=113nn2
Solution 1.
This series is similar to the one in Example 9.3.9, but now we are considering “3nn2” instead of “3n+n2.” This difference makes applying the Direct Comparison Test difficult.
Instead, we use the Limit Comparison Test and compare with the series n=113n:
limn1/(3nn2)1/3n=limn3n3nn2=1 (after applying L'Hospital's Rule twice) .
We know n=113n is a convergent geometric series, hence n=113nn2 converges as well.
Solution 2. Video solution
As mentioned before, practice helps one develop the intuition to quickly choose a series with which to compare. A general rule of thumb is to pick a series based on the dominant term in the expression of {an}. It is also helpful to note that factorials dominate exponentials, which dominate algebraic functions (e.g., polynomials), which dominate logarithms. In the previous example, the dominant term of 13nn2 was 3n, so we compared the series to n=113n. It is hard to apply the Limit Comparison Test to series containing factorials, though, as we have not learned how to apply L’Hospital’s Rule to n!.

Example 9.3.16. Applying the Limit Comparison Test.

Determine the convergence of n=1n+3n2n+1.
Solution 1.
We naïvely attempt to apply the rule of thumb given above and note that the dominant term in the expression of the series is 1/n2. Knowing that n=11n2 converges, we attempt to apply the Limit Comparison Test:
limn(n+3)/(n2n+1)1/n2=limnn2(n+3)n2n+1= (Apply L'Hospital's Rule) .
Theorem 9.3.12 part (3) only applies when n=1bn diverges; in our case, it converges. Ultimately, our test has not revealed anything about the convergence of our series.
The problem is that we chose a poor series with which to compare. Since the numerator and denominator of the terms of the series are both algebraic functions, we should have compared our series to the dominant term of the numerator divided by the dominant term of the denominator.
The dominant term of the numerator is n1/2 and the dominant term of the denominator is n2. Thus we should compare the terms of the given series to n1/2/n2=1/n3/2:
limn(n+3)/(n2n+1)1/n3/2=limnn3/2(n+3)n2n+1=1 (Apply L'Hospital's Rule) .
Since the p-series n=11n3/2 converges, we conclude that n=1n+3n2n+1 converges as well.
Solution 2. Video solution
We mentioned earlier that the Integral Test did not work well with series containing factorial terms. The next section introduces the Ratio Test, which does handle such series well. We also introduce the Root Test, which is good for series where each term is raised to a power.

Exercises 9.3.4 Exercises

Terms and Concepts

1.
In order to apply the Integral Test to a sequence {an}, the function a(n)=an must be ,  and .
2.
T/F: The Integral Test can be used to determine the sum of a convergent series.
3.
What test(s) in this section do not work well with factorials?
4.
Suppose n=0an is convergent, and there are sequences {bn} and {cn} such that bnancn for all n. What can be said about the series n=0bn and n=0cn?

Problems

Exercise Group.
In the following exercises, use the Integral Test to determine the convergence of the given series.
Exercise Group.
In the following exercises, use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison.
Exercise Group.
In the following exercises, use the Limit Comparison Test to determine the convergence of the given series; state what series is used for comparison.
Exercise Group.
In the following exercises, determine the convergence of the given series. State the test used; more than one test may be appropriate.
41.
Given that n=1an converges, state which of the following series converges, may converge, or does not converge.
42.
In this exercise, we explore an approximation method for series to which the Integral Test applies.
(a)
Let a(x) be a function to which the Integral Test applies, and for which the series n=1an converges.
Let Rn=n+1an denote the remainder; that is, the difference between n=1an and the nth partial sum. (Note that Rn is the size of the error that results if we approximate the series by the nth partial sum.) Explain why we must have the following inequality:
na(x)dxRnn+1a(x)dx
(b)
Estimate the error involved in using the first 12 terms to approximate the series n=11/n4. What is the approximate value of the series?
(c)
How many terms must we take to ensure that the nth partial sum approximation for n=11/n4 is accurate to 5 decimal places?
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