Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

9.4: Comparison Tests

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

( \newcommand{\kernel}{\mathrm{null}\,}\)

Learning Objectives
  • Use the comparison test to test a series for convergence.
  • Use the limit comparison test to determine convergence of a series.

We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. In this section, we show how to use comparison tests to determine the convergence or divergence of a series by comparing it to a series whose convergence or divergence is known. Typically these tests are used to determine convergence of series that are similar to geometric series or p-series.

Comparison Test

In the preceding two sections, we discussed two large classes of series: geometric series and p-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the comparison test.

For example, consider the series

n=11n2+1.

This series looks similar to the convergent series

n=11n2

Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing. Furthermore, since

0<1n2+1<1n2

for all positive integers n, the kth partial sum Sk of n=11n2+1 satisfies

Sk=kn=11n2+1<kn=11n2<n=11n2.

(See Figure 9.4.1a and Table 9.4.1.) Since the series on the right converges, the sequence Sk is bounded above. We conclude that Sk is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, Sk converges, and thus

n=11n2+1

converges.

Similarly, consider the series

n=11n1/2.

This series looks similar to the divergent series

n=11n.

The sequence of partial sums for each series is monotone increasing and

1n1/2>1n>0

for every positive integer n. Therefore, the kth partial sum Sk of

n=11n1/2

satisfies

Sk=kn=11n1/2>kn=11n.

(See Figure 9.4.1n and Table 9.4.1). Since the series n=11n diverges to infinity, the sequence of partial sums kn=11n is unbounded. Consequently, Sk is an unbounded sequence, and therefore diverges. We conclude that

n=11n1/2

diverges.

This shows two graphs side by side. The first shows plotted points for the partial sums for the sum of 1/n^2 and the sum 1/(n^2 + 1). Each of the partial sums for the latter is less than the corresponding partial sum for the former. The second shows plotted points for the partial sums for the sum of 1/(n - 0.5) and the sum 1/n. Each of the partial sums for the latter is less than the corresponding partial sum for the former.
Figure 9.4.1: (a) Each of the partial sums for the given series is less than the corresponding partial sum for the converging pseries. (b) Each of the partial sums for the given series is greater than the corresponding partial sum for the diverging harmonic series.
Table 9.4.1: Comparing a series with a p-series (p=2)
k 1 2 3 4 5 6 7 8
kn=11n2+1 0.5 0.7 0.8 0.8588 0.8973 0.9243 0.9443 0.9597
kn=11n2 1 1.25 1.3611 1.4236 1.4636 1.4914 1.5118 1.5274
Table 9.4.2: Comparing a series with the harmonic series
k 1 2 3 4 5 6 7 8
kn=11n1/2 2 2.6667 3.0667 3.3524 3.5746 3.7564 3.9103 4.0436
kn=11n 1 1.5 1.8333 2.0933 2.2833 2.45 2.5929 2.7179
Comparison Test
  1. Suppose there exists an integer N such that 0anbn for all nN. If n=1bn converges, then n=1an converges.
  2. Suppose there exists an integer N such that anbn0 for all nN. If n=1bn diverges, then n=1an diverges.
Proof

We prove part i. The proof of part ii. is the contrapositive of part i. Let Sk be the sequence of partial sums associated with n=1an, and let L=n=1bn. Since the terms an0,

Sk=a1+a2++aka1+a2++ak+ak+1=Sk+1.

Therefore, the sequence of partial sums is increasing. Further, since anbn for all nN, then

kn=Nankn=Nbnn=1bn=L.

Therefore, for all k1,

Sk=(a1+a2++aN1)+kn=Nan(a1+a2++aN1)+L.

Since a1+a2++aN1 is a finite number, we conclude that the sequence Sk is bounded above. Therefore, Sk is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that Sk converges, and therefore the series n=1an converges.

To use the comparison test to determine the convergence or divergence of a series n=1an, it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and p-series, these series are often used. If there exists an integer N such that for all nN, each term an is less than each corresponding term of a known convergent series, then n=1an converges. Similarly, if there exists an integer N such that for all nN, each term an is greater than each corresponding term of a known divergent series, then n=1an diverges.

Example 9.4.1: Using the Comparison Test

For each of the following series, use the comparison test to determine whether the series converges or diverges.

  1. n=11n3+3n+1
  2. n=112n+1
  3. n=21lnn
Solution

a. Compare to n=11n3. Since n=11n3 is a p-series with p=3, it converges. Further,

1n3+3n+1<1n3

for every positive integer n. Therefore, we can conclude that n=11n3+3n+1 converges.

b. Compare to n=1(12)n. Since n=1(12)n is a geometric series with r=12 and |12|<1, it converges. Also,

12n+1<12n

for every positive integer n. Therefore, we see that n=112n+1 converges.

c. Compare to n=21n. Since

1lnn>1n

for every integer n2 and n=21n diverges, we have that n=21lnn diverges.

Exercise 9.4.1

Use the comparison test to determine if the series n=1nn3+n+1 converges or diverges.

Hint

Find a value p such that nn3+n+11np.

Answer

The series converges.

Limit Comparison Test

The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series

n=21n21.

It is natural to compare this series with the convergent series

n=21n2.

However, this series does not satisfy the hypothesis necessary to use the comparison test because

1n21>1n2

for all integers n2. Although we could look for a different series with which to compare n=21n21, instead we show how we can use the limit comparison test to compare

n=21n21

and

n=21n2.

Let us examine the idea behind the limit comparison test. Consider two series n=1an and n=1bn. with positive terms an and bn and evaluate

limnanbn.

If

limnanbn=L0,

then, for n sufficiently large, anLbn. Therefore, either both series converge or both series diverge. For the series n=21n21 and n=21n2, we see that

limn1/(n21)1/n2=limnn2n21=1.

Since n=21n2 converges, we conclude that

n=21n21

converges.

The limit comparison test can be used in two other cases. Suppose

limnanbn=0.

In this case, an/bn is a bounded sequence. As a result, there exists a constant M such that anMbn. Therefore, if n=1bn converges, then n=1an converges. On the other hand, suppose

limnanbn=.

In this case, an/bn is an unbounded sequence. Therefore, for every constant M there exists an integer N such that anMbn for all nN. Therefore, if n=1bn diverges, then n=1an diverges as well.

Limit Comparison Test

Let an,bn0 for all n1.

  1. If limnanbn=L0, then n=1an and n=1bn both converge or both diverge.
  2. If limnanbn=0 and n=1bn converges, then n=1an converges.
  3. If limnanbn= and n=1bn diverges, then n=1an diverges.

Note that if anbn0 and n=1bn diverges, the limit comparison test gives no information. Similarly, if anbn and n=1bn converges, the test also provides no information. For example, consider the two series n=11n and n=11n2. These series are both p-series with p=12 and p=2, respectively. Since p=12<1, the series n=11n diverges. On the other hand, since p=2>1, the series n=11n2 converges. However, suppose we attempted to apply the limit comparison test, using the convergent p−series n=11n3 as our comparison series. First, we see that

1/n1/n3=n3n=n5/2 as n.

Similarly, we see that

1/n21/n3=n as n.

Therefore, if anbn when n=1bn converges, we do not gain any information on the convergence or divergence of n=1an.

Example 9.4.2: Using the Limit Comparison Test

For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.

  1. n=11n+1
  2. n=12n+13n
  3. n=1ln(n)n2
Solution

a. Compare this series to n=11n. Calculate

limn1/(n+1)1/n=limnnn+1=limn1/n1+1/n=1.

By the limit comparison test, since n=11n diverges, then n=11n+1 diverges.

b. Compare this series to n=1(23)n. We see that

limn(2n+1)/3n2n/3n=limn2n+13n3n2n=limn2n+12n=limn[1+(12)n]=1.

Therefore,

limn(2n+1)/3n2n/3n=1.

Since n=1(23)n converges, we conclude that n=12n+13n converges.

c. Since lnn<n, compare with n=11n. We see that

limnlnn/n21/n=limnlnnn2n1=limnlnnn.

In order to evaluate limnlnn/n, evaluate the limit as x of the real-valued function ln(x)/x. These two limits are equal, and making this change allows us to use L’Hôpital’s rule. We obtain

limxlnxx=limx1x=0.

Therefore, limnlnnn=0, and, consequently,

limn(lnn)/n21/n=0.

Since the limit is 0 but n=11n diverges, the limit comparison test does not provide any information.

Compare with n=11n2 instead. In this case,

limn(lnn)/n21/n2=limnlnnn2n21=limnlnn=.

Since the limit is but n=11n2 converges, the test still does not provide any information.

So now we try a series between the two we already tried. Choosing the series n=11n3/2, we see that

limn(lnn)/n21/n3/2=limnlnnn2n3/21=limnlnnn.

As above, in order to evaluate limnlnnn, evaluate the limit as x of the real-valued function lnnn. Using L’Hôpital’s rule,

limxlnxx=limx2xx=limx2x=0.

Since the limit is 0 and n=11n3/2 converges, we can conclude that n=1lnnn2 converges.

Exercise 9.4.2

Use the limit comparison test to determine whether the series n=15n3n+2 converges or diverges.

Hint

Compare with a geometric series.

Answer

The series diverges.

Key Concepts

  • The comparison tests are used to determine convergence or divergence of series with positive terms.
  • When using the comparison tests, a series n=1an is often compared to a geometric or p-series.

Glossary

comparison test
If 0anbn for all nN and n=1bn converges, then n=1an converges; if anbn0 for all nN and n=1bn diverges, then n=1an diverges.
limit comparison test
Suppose an,bn0 for all n1. If limnan/bnL0, then n=1an and n=1bn both converge or both diverge; if limnan/bn0 and n=1bn converges, then n=1an converges. If limnan/bn, and n=1bn diverges, then n=1an diverges.

This page titled 9.4: Comparison Tests is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?