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4.4: Comparison Tests

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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

\sum_{n=1}^∞\dfrac{1}{n^2+1}. \nonumber

This series looks similar to the convergent series

\sum_{n=1}^∞\dfrac{1}{n^2} \nonumber

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

0<\dfrac{1}{n^2+1}<\dfrac{1}{n^2} \nonumber

for all positive integers n, the k^{\text{th}} partial sum S_k of \displaystyle \sum^∞_{n=1}\dfrac{1}{n^2+1} satisfies

S_k=\sum_{n=1}^k\dfrac{1}{n^2+1}<\sum_{n=1}^k\dfrac{1}{n^2}<\sum_{n=1}^∞\dfrac{1}{n^2}. \nonumber

(See Figure \PageIndex{1a} and Table \PageIndex{1}.) Since the series on the right converges, the sequence {S_k} is bounded above. We conclude that {S_k} is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, {S_k} converges, and thus

\sum_{n=1}^∞\dfrac{1}{n^2+1} \nonumber

converges.

Similarly, consider the series

\sum_{n=1}^∞\dfrac{1}{n−1/2}. \nonumber

This series looks similar to the divergent series

\sum_{n=1}^∞\dfrac{1}{n}. \nonumber

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

\dfrac{1}{n−1/2}>\dfrac{1}{n}>0 \nonumber

for every positive integer n. Therefore, the k^{\text{th}} partial sum S_k of

\sum^∞_{n=1}\dfrac{1}{n−1/2} \nonumber

satisfies

S_k=\sum_{n=1}^k\dfrac{1}{n−1/2}>\sum_{n=1}^k\dfrac{1}{n}. \nonumber

(See Figure \PageIndex{1n} and Table \PageIndex{1}). Since the series \displaystyle \sum^∞_{n=1}\frac{1}{n} diverges to infinity, the sequence of partial sums \displaystyle \sum^k_{n=1}\frac{1}{n} is unbounded. Consequently, {S_k} is an unbounded sequence, and therefore diverges. We conclude that

\sum_{n=1}^∞\dfrac{1}{n−1/2} \nonumber

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 \PageIndex{1}: (a) Each of the partial sums for the given series is less than the corresponding partial sum for the converging p−series. (b) Each of the partial sums for the given series is greater than the corresponding partial sum for the diverging harmonic series.
Table \PageIndex{1}: Comparing a series with a p-series (p = 2)
k 1 2 3 4 5 6 7 8
\displaystyle \sum_{n=1}^k\dfrac{1}{n^2+1} 0.5 0.7 0.8 0.8588 0.8973 0.9243 0.9443 0.9597
\displaystyle \sum_{n=1}^k\dfrac{1}{n^2} 1 1.25 1.3611 1.4236 1.4636 1.4914 1.5118 1.5274
Table \PageIndex{2}: Comparing a series with the harmonic series
k 1 2 3 4 5 6 7 8
\displaystyle \sum_{n=1}^k\dfrac{1}{n−1/2} 2 2.6667 3.0667 3.3524 3.5746 3.7564 3.9103 4.0436
\displaystyle \sum_{n=1}^k\dfrac{1}{n} 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 0≤a_n≤b_n for all n≥N. If \displaystyle \sum^∞_{n=1}b_n converges, then \displaystyle \sum^∞_{n=1}a_n converges.
  2. Suppose there exists an integer N such that a_n≥b_n≥0 for all n≥N. If \displaystyle \sum^∞_{n=1}b_n diverges, then \displaystyle \sum^∞_{n=1}a_n diverges.
Proof

We prove part i. The proof of part ii. is the contrapositive of part i. Let {S_k} be the sequence of partial sums associated with \displaystyle \sum^∞_{n=1}a_n, and let \displaystyle L=\sum^∞_{n=1}b_n. Since the terms a_n≥0,

S_k=a_1+a_2+⋯+a_k≤a_1+a_2+⋯+a_k+a_{k+1}=S_{k+1}. \nonumber

Therefore, the sequence of partial sums is increasing. Further, since a_n≤b_n for all n≥N, then

\sum_{n=N}^ka_n≤\sum_{n=N}^kb_n≤\sum_{n=1}^∞b_n=L. \nonumber

Therefore, for all k≥1,

S_k=(a_1+a_2+⋯+a_{N−1})+\sum_{n=N}^ka_n≤(a_1+a_2+⋯+a_{N−1})+L. \nonumber

Since a_1+a_2+⋯+a_{N−1} is a finite number, we conclude that the sequence {S_k} is bounded above. Therefore, {S_k} is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that {S_k} converges, and therefore the series \displaystyle \sum_{n=1}^∞a_n converges.

To use the comparison test to determine the convergence or divergence of a series \displaystyle \sum_{n=1}^∞a_n, 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 \displaystyle \sum_{n=1}^∞a_n 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 \displaystyle \sum_{n=1}^∞a_n diverges.

Example \PageIndex{1}: Using the Comparison Test

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

  1. \displaystyle \sum_{n=1}^∞\dfrac{1}{n^3+3n+1}
  2. \displaystyle \sum_{n=1}^∞\dfrac{1}{2^n+1}
  3. \displaystyle \sum_{n=2}^∞\dfrac{1}{\ln \,n }
Solution

a. Compare to \displaystyle \sum_{n=1}^∞\dfrac{1}{n^3}. Since \displaystyle \sum_{n=1}^∞\dfrac{1}{n^3} is a p-series with p=3, it converges. Further,

\dfrac{1}{n^3+3n+1}<\dfrac{1}{n^3} \nonumber

for every positive integer n. Therefore, we can conclude that \displaystyle \sum^∞_{n=1}\dfrac{1}{n^3+3n+1} converges.

b. Compare to \displaystyle \sum^∞_{n=1}\left(\dfrac{1}{2}\right)^n. Since \displaystyle \sum_{n=1}^∞\left(\dfrac{1}{2}\right)^n is a geometric series with r=\dfrac{1}{2} and \left|\dfrac{1}{2}\right|<1, it converges. Also,

\dfrac{1}{2^n+1}<\dfrac{1}{2^n} \nonumber

for every positive integer n. Therefore, we see that \displaystyle \sum^∞_{n=1}\dfrac{1}{2^n+1} converges.

c. Compare to \displaystyle \sum^∞_{n=2}\dfrac{1}{n}. Since

\dfrac{1}{\ln n }>\dfrac{1}{n} \nonumber

for every integer n≥2 and \displaystyle \sum^∞_{n=2}\dfrac{1}{n} diverges, we have that \displaystyle \sum^∞_{n=2}\dfrac{1}{\ln n} diverges.

Exercise \PageIndex{1}

Use the comparison test to determine if the series \displaystyle \sum^∞_{n=1}\dfrac{n}{n^3+n+1} converges or diverges.

Hint

Find a value p such that \dfrac{n}{n^3+n+1}≤\dfrac{1}{n^p}.

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

\sum_{n=2}^∞\dfrac{1}{n^2−1}. \nonumber

It is natural to compare this series with the convergent series

\sum_{n=2}^∞\dfrac{1}{n^2}. \nonumber

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

\dfrac{1}{n^2−1}>\dfrac{1}{n^2} \nonumber

for all integers n≥2. Although we could look for a different series with which to compare \displaystyle \sum^∞_{n=2}\frac{1}{n^2−1}, instead we show how we can use the limit comparison test to compare

\sum_{n=2}^∞\frac{1}{n^2−1} \nonumber

and

\sum_{n=2}^∞\frac{1}{n^2}. \nonumber

Let us examine the idea behind the limit comparison test. Consider two series \displaystyle \sum^∞_{n=1}a_n and \displaystyle \sum^∞_{n=1}b_n. with positive terms a_n and b_n and evaluate

\lim_{n→∞}\frac{a_n}{b_n}. \nonumber

If

\lim_{n→∞}\frac{a_n}{b_n}=L≠0, \nonumber

then, for n sufficiently large, a_n≈Lb_n. Therefore, either both series converge or both series diverge. For the series \displaystyle \sum^∞_{n=2}\frac{1}{n^2−1} and \displaystyle \sum^∞_{n=2}\dfrac{1}{n^2}, we see that

\lim_{n→∞}\dfrac{1/(n^2−1)}{1/n^2}=\lim_{n→∞}\dfrac{n^2}{n^2−1}=1. \nonumber

Since \displaystyle \sum^∞_{n=2}\frac{1}{n^2} converges, we conclude that

\sum_{n=2}^∞\dfrac{1}{n^2−1} \nonumber

converges.

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

\lim_{n→∞}\dfrac{a_n}{b_n}=0. \nonumber

In this case, {a_n/b_n} is a bounded sequence. As a result, there exists a constant M such that a_n≤Mb_n. Therefore, if \displaystyle \sum^∞_{n=1}b_n converges, then \displaystyle \sum^∞_{n=1}a_n converges. On the other hand, suppose

\lim_{n→∞}\dfrac{a_n}{b_n}=∞. \nonumber

In this case, {a_n/b_n} is an unbounded sequence. Therefore, for every constant M there exists an integer N such that a_n≥Mb_n for all n≥N. Therefore, if \displaystyle \sum^∞_{n=1}b_n diverges, then \displaystyle \sum^∞_{n=1}a_n diverges as well.

Limit Comparison Test

Let a_n,b_n≥0 for all n≥1.

  1. If \displaystyle \lim_{n→∞}\frac{a_n}{b_n}=L≠0, then \displaystyle \sum^∞_{n=1}a_n and \displaystyle \sum^∞_{n=1}b_n both converge or both diverge.
  2. If \displaystyle \lim_{n→∞}\frac{a_n}{b_n}=0 and \displaystyle \sum^∞_{n=1}b_n converges, then \displaystyle \sum^∞_{n=1}a_n converges.
  3. If \displaystyle \lim_{n→∞}\frac{a_n}{b_n}=∞ and \displaystyle \sum^∞_{n=1}b_n diverges, then \displaystyle \sum^∞_{n=1}a_n diverges.

Note that if \dfrac{a_n}{b_n}→0 and \displaystyle \sum^∞_{n=1}b_n diverges, the limit comparison test gives no information. Similarly, if \dfrac{a_n}{b_n}→∞ and \displaystyle \sum^∞_{n=1}b_n converges, the test also provides no information. For example, consider the two series \displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n}} and \displaystyle \sum_{n=1}^∞\frac{1}{n^2}. These series are both p-series with p=\frac{1}{2} and p=2, respectively. Since p=\frac{1}{2}<1, the series \displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n}} diverges. On the other hand, since p=2>1, the series \displaystyle \sum_{n=1}^∞\frac{1}{n^2} converges. However, suppose we attempted to apply the limit comparison test, using the convergent p−series \displaystyle \sum_{n=1}^∞\frac{1}{n^3} as our comparison series. First, we see that

\dfrac{1/\sqrt{n}}{1/n^3}=\dfrac{n^3}{\sqrt{n}}=n^{5/2}→∞\; \text{ as } \;n→∞. \nonumber

Similarly, we see that

\dfrac{1/n^2}{1/n^3}=n→∞\; \text{ as } \;n→∞. \nonumber

Therefore, if \dfrac{a_n}{b_n}→∞ when \displaystyle \sum_{n=1}^∞b_n converges, we do not gain any information on the convergence or divergence of \displaystyle \sum_{n=1}^∞a_n.

Example \PageIndex{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. \displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}+1}
  2. \displaystyle \sum^∞_{n=1}\dfrac{2^n+1}{3^n}
  3. \displaystyle \sum^∞_{n=1}\dfrac{\ln(n)}{n^2}
Solution

a. Compare this series to \displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}}. Calculate

\displaystyle \lim_{n→∞}\dfrac{1/(\sqrt{n}+1)}{1/\sqrt{n}}=\lim_{n→∞}\dfrac{\sqrt{n}}{\sqrt{n}+1}=\lim_{n→∞}\dfrac{1/\sqrt{n}}{1+1/\sqrt{n}}=1.

By the limit comparison test, since \displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}} diverges, then \displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}+1} diverges.

b. Compare this series to \displaystyle \sum^∞_{n=1}\left(\dfrac{2}{3}\right)^n. We see that

\displaystyle \lim_{n→∞}\dfrac{(2^n+1)/3^n}{2^n/3^n}=\lim_{n→∞}\dfrac{2^n+1}{3^n}⋅\dfrac{3^n}{2^n}=\lim_{n→∞}\dfrac{2^n+1}{2^n}=\lim_{n→∞}\left[1+\left(\tfrac{1}{2}\right)^n\right]=1.

Therefore,

\displaystyle \lim_{n→∞}\dfrac{(2^n+1)/3^n}{2^n/3^n}=1.

Since \displaystyle \sum^∞_{n=1}\left(\dfrac{2}{3}\right)^n converges, we conclude that \displaystyle \sum^∞_{n=1}\dfrac{2^n+1}{3^n} converges.

c. Since \ln n<n, compare with \displaystyle \sum^∞_{n=1}\dfrac{1}{n}. We see that

\displaystyle \lim_{n→∞}\dfrac{\ln n/n^2}{1/n}=\lim_{n→∞}\dfrac{\ln n}{n^2}⋅\dfrac{n}{1}=\lim_{n→∞}\dfrac{\ln n}{n}.

In order to evaluate \displaystyle \lim_{n→∞}\ln n/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

\displaystyle \lim_{x→∞}\dfrac{lnx}{x}=\lim_{x→∞}\dfrac{1}{x}=0.

Therefore, \displaystyle \lim_{n→∞}\frac{\ln n}{n}=0, and, consequently,

\displaystyle \lim_{n→∞}\dfrac{(\ln n)/n^2}{1/n}=0.

Since the limit is 0 but \displaystyle \sum^∞_{n=1}\dfrac{1}{n} diverges, the limit comparison test does not provide any information.

Compare with \displaystyle \sum^∞_{n=1}\dfrac{1}{n^2} instead. In this case,

\displaystyle \lim_{n→∞}\dfrac{(\ln n)/n^2}{1/n^2}=\lim_{n→∞}\dfrac{\ln n}{n^2}⋅\dfrac{n^2}{1}=\lim_{n→∞}\ln n=∞.

Since the limit is but \displaystyle \sum^∞_{n=1}\dfrac{1}{n^2} converges, the test still does not provide any information.

So now we try a series between the two we already tried. Choosing the series \displaystyle \sum^∞_{n=1}\dfrac{1}{n^{3/2}}, we see that

\displaystyle \lim_{n→∞}\dfrac{(\ln n)/n^2}{1/n^{3/2}}=\lim_{n→∞}\dfrac{\ln n}{n^2}⋅\dfrac{n^{3/2}}{1}=\lim_{n→∞}\dfrac{\ln n}{\sqrt{n}}.

As above, in order to evaluate \displaystyle \lim_{n→∞}\frac{\ln n}{\sqrt{n}}, evaluate the limit as x→∞ of the real-valued function \frac{\ln n}{\sqrt{n}}. Using L’Hôpital’s rule,

\displaystyle \lim_{x→∞}\dfrac{\ln x}{\sqrt{x}}=\lim_{x→∞}\dfrac{2\sqrt{x}}{x}=\lim_{x→∞}\dfrac{2}{\sqrt{x}}=0.

Since the limit is 0 and \displaystyle \sum^∞_{n=1}\dfrac{1}{n^{3/2}} converges, we can conclude that \displaystyle \sum^∞_{n=1}\dfrac{\ln n}{n^2} converges.

Exercise \PageIndex{2}

Use the limit comparison test to determine whether the series \displaystyle \sum^∞_{n=1}\dfrac{5^n}{3^n+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 \displaystyle \sum^∞_{n=1}a_n is often compared to a geometric or p-series.

Glossary

comparison test
If 0≤a_n≤b_n for all n≥N and \displaystyle \sum^∞_{n=1}b_n converges, then \displaystyle \sum^∞_{n=1}a_n converges; if a_n≥b_n≥0 for all n≥N and \displaystyle \sum^∞_{n=1}b_n diverges, then \displaystyle \sum^∞_{n=1}a_n diverges.
limit comparison test
Suppose a_n,b_n≥0 for all n≥1. If \displaystyle \lim_{n→∞}a_n/b_n→L≠0, then \displaystyle \sum^∞_{n=1}a_n and \displaystyle \sum^∞_{n=1}b_n both converge or both diverge; if \displaystyle \lim_{n→∞}a_n/b_n→0 and \displaystyle \sum^∞_{n=1}b_n converges, then \displaystyle \sum^∞_{n=1}a_n converges. If \displaystyle \lim_{n→∞}a_n/b_n→∞, and \displaystyle \sum^∞_{n=1}b_n diverges, then \displaystyle \sum^∞_{n=1}a_n diverges.

This page titled 4.4: Comparison Tests is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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