9.4: Comparison Tests
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- 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=k∑n=11n2+1<k∑n=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=11n−1/2.
This series looks similar to the divergent series
∞∑n=11n.
The sequence of partial sums for each series is monotone increasing and
1n−1/2>1n>0
for every positive integer n. Therefore, the kth partial sum Sk of
∞∑n=11n−1/2
satisfies
Sk=k∑n=11n−1/2>k∑n=11n.
(See Figure 9.4.1n and Table 9.4.1). Since the series ∞∑n=11n diverges to infinity, the sequence of partial sums k∑n=11n is unbounded. Consequently, Sk is an unbounded sequence, and therefore diverges. We conclude that
∞∑n=11n−1/2
diverges.
k | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
---|---|---|---|---|---|---|---|---|
k∑n=11n2+1 | 0.5 | 0.7 | 0.8 | 0.8588 | 0.8973 | 0.9243 | 0.9443 | 0.9597 |
k∑n=11n2 | 1 | 1.25 | 1.3611 | 1.4236 | 1.4636 | 1.4914 | 1.5118 | 1.5274 |
k | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
---|---|---|---|---|---|---|---|---|
k∑n=11n−1/2 | 2 | 2.6667 | 3.0667 | 3.3524 | 3.5746 | 3.7564 | 3.9103 | 4.0436 |
k∑n=11n | 1 | 1.5 | 1.8333 | 2.0933 | 2.2833 | 2.45 | 2.5929 | 2.7179 |
- Suppose there exists an integer N such that 0≤an≤bn for all n≥N. If ∞∑n=1bn converges, then ∞∑n=1an converges.
- Suppose there exists an integer N such that an≥bn≥0 for all n≥N. If ∞∑n=1bn diverges, then ∞∑n=1an diverges.
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 an≥0,
Sk=a1+a2+⋯+ak≤a1+a2+⋯+ak+ak+1=Sk+1.
Therefore, the sequence of partial sums is increasing. Further, since an≤bn for all n≥N, then
k∑n=Nan≤k∑n=Nbn≤∞∑n=1bn=L.
Therefore, for all k≥1,
Sk=(a1+a2+⋯+aN−1)+k∑n=Nan≤(a1+a2+⋯+aN−1)+L.
Since a1+a2+⋯+aN−1 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 n≥N, 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 n≥N, each term an is greater than each corresponding term of a known divergent series, then ∞∑n=1an diverges.
For each of the following series, use the comparison test to determine whether the series converges or diverges.
- ∞∑n=11n3+3n+1
- ∞∑n=112n+1
- ∞∑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 n≥2 and ∞∑n=21n diverges, we have that ∞∑n=21lnn diverges.
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+1≤1np.
- 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=21n2−1.
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
1n2−1>1n2
for all integers n≥2. Although we could look for a different series with which to compare ∞∑n=21n2−1, instead we show how we can use the limit comparison test to compare
∞∑n=21n2−1
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
limn→∞anbn.
If
limn→∞anbn=L≠0,
then, for n sufficiently large, an≈Lbn. Therefore, either both series converge or both series diverge. For the series ∞∑n=21n2−1 and ∞∑n=21n2, we see that
limn→∞1/(n2−1)1/n2=limn→∞n2n2−1=1.
Since ∞∑n=21n2 converges, we conclude that
∞∑n=21n2−1
converges.
The limit comparison test can be used in two other cases. Suppose
limn→∞anbn=0.
In this case, an/bn is a bounded sequence. As a result, there exists a constant M such that an≤Mbn. Therefore, if ∞∑n=1bn converges, then ∞∑n=1an converges. On the other hand, suppose
limn→∞anbn=∞.
In this case, an/bn is an unbounded sequence. Therefore, for every constant M there exists an integer N such that an≥Mbn for all n≥N. Therefore, if ∞∑n=1bn diverges, then ∞∑n=1an diverges as well.
Let an,bn≥0 for all n≥1.
- If limn→∞anbn=L≠0, then ∞∑n=1an and ∞∑n=1bn both converge or both diverge.
- If limn→∞anbn=0 and ∞∑n=1bn converges, then ∞∑n=1an converges.
- If limn→∞anbn=∞ and ∞∑n=1bn diverges, then ∞∑n=1an diverges.
Note that if anbn→0 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=11√n and ∞∑n=11n2. These series are both p-series with p=12 and p=2, respectively. Since p=12<1, the series ∞∑n=11√n 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=n3√n=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.
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.
- ∞∑n=11√n+1
- ∞∑n=12n+13n
- ∞∑n=1ln(n)n2
Solution
a. Compare this series to ∞∑n=11√n. Calculate
limn→∞1/(√n+1)1/√n=limn→∞√n√n+1=limn→∞1/√n1+1/√n=1.
By the limit comparison test, since ∞∑n=11√n diverges, then ∞∑n=11√n+1 diverges.
b. Compare this series to ∞∑n=1(23)n. We see that
limn→∞(2n+1)/3n2n/3n=limn→∞2n+13n⋅3n2n=limn→∞2n+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
limn→∞lnn/n21/n=limn→∞lnnn2⋅n1=limn→∞lnnn.
In order to evaluate limn→∞lnn/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
limx→∞lnxx=limx→∞1x=0.
Therefore, limn→∞lnnn=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=limn→∞lnnn2⋅n21=limn→∞lnn=∞.
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=limn→∞lnnn2⋅n3/21=limn→∞lnn√n.
As above, in order to evaluate limn→∞lnn√n, evaluate the limit as x→∞ of the real-valued function lnn√n. Using L’Hôpital’s rule,
limx→∞lnx√x=limx→∞2√xx=limx→∞2√x=0.
Since the limit is 0 and ∞∑n=11n3/2 converges, we can conclude that ∞∑n=1lnnn2 converges.
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 0≤an≤bn for all n≥N and ∞∑n=1bn converges, then ∞∑n=1an converges; if an≥bn≥0 for all n≥N and ∞∑n=1bn diverges, then ∞∑n=1an diverges.
- limit comparison test
- Suppose an,bn≥0 for all n≥1. If limn→∞an/bn→L≠0, then ∞∑n=1an and ∞∑n=1bn both converge or both diverge; if limn→∞an/bn→0 and ∞∑n=1bn converges, then ∞∑n=1an converges. If limn→∞an/bn→∞, and ∞∑n=1bn diverges, then ∞∑n=1an diverges.