3.4: Comparison Tests

Learning Objectives

• Use the Series 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.

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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 Series Comparison Test.

For example, consider the series

\[\sum_{n=1}^ \infty \dfrac{1}{n^2+1}. \nonumber \]

This series looks similar to the convergent series

\[\sum_{n=1}^ \infty \dfrac{1}{n^2} \nonumber \]

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

\[0 \lt \dfrac{1}{n^2+1} \lt \dfrac{1}{n^2} \nonumber \]

for all \(n \in \mathbb{N}\), the \(k^{\text{th}}\) partial sum \(S_k\) of \(\displaystyle \sum^ \infty _{n=1}\dfrac{1}{n^2+1}\) satisfies

\[S_k=\sum_{n=1}^k\dfrac{1}{n^2+1} \lt \sum_{n=1}^k\dfrac{1}{n^2} \lt \sum_{n=1}^ \infty \dfrac{1}{n^2}. \nonumber \]

(See Figure \(\PageIndex{1}a\) and Table \(\PageIndex{1}\) below.) 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

\[\sum_{n=1}^ \infty \dfrac{1}{n^2+1} \nonumber \]

converges.

Similarly, consider the series

\[\sum_{n=1}^ \infty \dfrac{1}{n−1/2}. \nonumber \]

This series looks similar to the divergent series

\[\sum_{n=1}^ \infty \dfrac{1}{n}. \nonumber \]

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

\[\dfrac{1}{n−1/2} \gt \dfrac{1}{n} \gt 0 \nonumber \]

for every \(n \in \mathbb{N}\). Therefore, the \(k^{\text{th}}\) partial sum \(S_k\) of

\[ \sum^ \infty _{n=1}\dfrac{1}{n−1/2} \nonumber \]

satisfies

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

(See Figure \(\PageIndex{1}b\) and Table \(\PageIndex{2}\) below). Since the series \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n}\) 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

\[\sum_{n=1}^ \infty \dfrac{1}{n−1/2} \nonumber \]

diverges.

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

Theorem: Series Comparison Test

1. Suppose there exists an integer \(N\) such that \(0 \leq a_n \leq b_n\) for all \(n \geq N\). If \(\displaystyle \sum^ \infty _{n=1}b_n\) converges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) converges.
2. Suppose there exists an integer \(N\) such that \(a_n \geq b_n \geq 0\) for all \(n \geq N.\) If \(\displaystyle \sum^ \infty _{n=1}b_n\) diverges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) 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 \(\displaystyle \sum^ \infty _{n=1}a_n\), and let \(\displaystyle L=\sum^ \infty _{n=1}b_n\). Since the terms \(a_n \geq 0\),\[S_k=a_1+a_2+ \cdots +a_k \leq a_1+a_2+ \cdots +a_k+a_{k+1}=S_{k+1}. \nonumber \]Therefore, the sequence of partial sums is increasing. Further, since \(a_n \leq b_n\) for all \(n \geq N\), \[\sum_{n=N}^ka_n \leq \sum_{n=N}^kb_n \leq \sum_{n=1}^ \infty b_n=L. \nonumber \]Therefore, for all \(k \geq 1\),\[S_k=(a_1+a_2+ \cdots +a_{N−1})+\sum_{n=N}^ka_n \leq (a_1+a_2+ \cdots +a_{N−1})+L. \nonumber \]Since \(a_1+a_2+ \cdots +a_{N−1}\) 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 \(\displaystyle \sum_{n=1}^ \infty a_n\) converges.

Caution

You must be certain that the terms of the original series and the series you are using to compare are positive! This is critical because the Series Comparison Test fails to work otherwise.

To use the Series Comparison Test to determine the convergence or divergence of a series \(\displaystyle \sum_{n=1}^ \infty a_n\), 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 \(n \geq N\), each term \(a_n\) is less than each corresponding term of a known convergent series, then \(\displaystyle \sum_{n=1}^ \infty a_n\) converges. Similarly, if there exists an integer \(N\) such that for all \(n \geq N\), each term \(a_n\) is greater than each corresponding term of a known divergent series, then \(\displaystyle \sum_{n=1}^ \infty a_n\) 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. \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^3+3n+1}\)
2. \(\displaystyle \sum_{n=1}^ \infty \frac{1}{2^n+1}\)
3. \(\displaystyle \sum_{n=2}^ \infty \frac{1}{\ln \,n }\)

Solution

a. Compare to \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^3}\). Note that the terms of the original series and the compared series are positive. Since \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^3}\) is a \(p\)-series with \(p=3\), it converges. Further,

\[n^3 + 3n + 1 \gt n^3 \gt 0 \implies 0 \lt \dfrac{1}{n^3+3n+1} \lt \dfrac{1}{n^3} \nonumber \]

for every \(n \in \mathbb{N}\). Therefore, we can conclude that \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n^3+3n+1}\) converges.

b. Compare to \(\displaystyle \sum^ \infty _{n=1}\left(\frac{1}{2}\right)^n\). Again, the terms of both series are positive. Since \(\displaystyle \sum_{n=1}^ \infty \left(\frac{1}{2}\right)^n\) is a geometric series with \(r=\frac{1}{2}\) and \(\left|\frac{1}{2}\right| \lt 1\), it converges. Also,

\[2^n + 1 \gt 2^n \gt 0 \implies 0 \lt \dfrac{1}{2^n+1} \lt \dfrac{1}{2^n} \nonumber \]

for every \(n \in \mathbb{N}\). Therefore, we see that \(\displaystyle \sum^ \infty _{n=1}\frac{1}{2^n+1}\) converges.

c. Compare to \(\displaystyle \sum^ \infty _{n=2}\frac{1}{n}\). The terms of both series are positive.

If we let \( f(x) = x \) and \( g(x) = \ln(x) \), we know that \( f(1) = 1 \gt 0 = g(1) \). Moreover, for \( x \geq 1 \), \( f^{\prime}(x) = 1 \geq g^{\prime}(x) = \frac{1}{x} \). Hence, by the Racetrack Principle (from Calculus I), \( f(x) = x \gt \ln(x) = g(x) \) for \( x \geq 1 \). Thus,

\[0 \lt \ln(n) \lt n \implies \dfrac{1}{\ln(n) } \gt \dfrac{1}{n} \gt 0 \nonumber \]

for \(n \geq 2\). Since \(\displaystyle \sum^ \infty _{n=2}\frac{1}{n}\) diverges, we have that \(\displaystyle \sum^ \infty _{n=2}\frac{1}{\ln n}\) diverges.

Exercise \(\PageIndex{1}\)

Use the Comparison Test to determine if the series \(\displaystyle \sum^ \infty _{n=1}\frac{n}{n^3+n+1}\) converges or diverges.

Hint

Find a value \(p\) such that \(\frac{n}{n^3+n+1} \leq \frac{1}{n^p}\).

Answer

The series converges.

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Limit Comparison Test

The Series 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

\[\sum_{n=2}^ \infty \dfrac{1}{n^2−1}. \nonumber \]

It is natural to compare this series with the convergent series

\[\sum_{n=2}^ \infty \dfrac{1}{n^2}. \nonumber \]

However, this series does not satisfy the hypothesis necessary to use the Series Comparison Test because

\[\dfrac{1}{n^2−1} \gt \dfrac{1}{n^2} \nonumber \]

for all integers \(n \geq 2\). Although we could look for a different series with which to compare \(\displaystyle \sum^ \infty _{n=2}\frac{1}{n^2−1},\) instead we show how we can use the Limit Comparison Test to compare

\[\sum_{n=2}^ \infty \dfrac{1}{n^2−1} \nonumber \]

and

\[\sum_{n=2}^ \infty \dfrac{1}{n^2}. \nonumber \]

Let us examine the idea behind the Limit Comparison Test. Consider two series \(\displaystyle \sum^ \infty _{n=1}a_n\) and \(\displaystyle \sum^ \infty _{n=1}b_n\) with positive terms \(a_n\) and \(b_n\) and evaluate

\[\lim_{n \to \infty }\frac{a_n}{b_n}. \nonumber \]

If

\[\lim_{n \to \infty }\frac{a_n}{b_n}=L \neq 0, \nonumber \]

then, for \(n\) sufficiently large, \(a_n \approx L b_n\). Therefore, either both series converge or both series diverge. For the series \(\displaystyle \sum^ \infty _{n=2}\frac{1}{n^2−1}\) and \(\displaystyle \sum^ \infty _{n=2}\dfrac{1}{n^2}\), we see that

\[\lim_{n \to \infty }\dfrac{1/(n^2−1)}{1/n^2}=\lim_{n \to \infty }\dfrac{n^2}{n^2−1}=1. \nonumber \]

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

\[\sum_{n=2}^ \infty \dfrac{1}{n^2−1} \nonumber \]

converges.

The Limit Comparison Test can be used in two other cases. Suppose

\[\lim_{n \to \infty }\dfrac{a_n}{b_n}=0. \nonumber \]

You can conceptually think of this as saying \( b_n\) will eventually be much greater than \( a_n \). In this case, \(\{a_n/b_n\}\) is a bounded sequence. As a result, there exists a constant \(M\) such that \(a_n \leq M b_n\). Therefore, if \(\displaystyle \sum^ \infty _{n=1}b_n\) converges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) converges. On the other hand, suppose

\[\lim_{n \to \infty }\dfrac{a_n}{b_n}= \infty . \nonumber \]

Again, conceptually this tells us that \( a_n \) is eventually much greater than \( b_n \). In this case, \(\{a_n/b_n\}\) is an unbounded sequence. Therefore, for every constant \(M\) there exists an integer \(N\) such that \(a_n \geq M b_n\) for all \(n \geq N\). Hence, if \(\displaystyle \sum^ \infty _{n=1}b_n\) diverges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) diverges as well.

We summarize these findings in the following theorem.

Theorem: Limit Comparison Test

Let \(a_n, b_n \geq 0\) for all \(n \geq 1\).

1. If \(\displaystyle \lim_{n \to \infty }\frac{a_n}{b_n}=L \neq 0,\) then \(\displaystyle \sum^ \infty _{n=1}a_n\) and \(\displaystyle \sum^ \infty _{n=1}b_n\) both converge or both diverge.
2. If \(\displaystyle \lim_{n \to \infty }\frac{a_n}{b_n}=0\) and \(\displaystyle \sum^ \infty _{n=1}b_n\) converges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) converges.
3. If \(\displaystyle \lim_{n \to \infty }\frac{a_n}{b_n}= \infty \) and \(\displaystyle \sum^ \infty _{n=1}b_n\) diverges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) diverges.

Note that if \(\frac{a_n}{b_n} \to 0\) and \(\displaystyle \sum^ \infty _{n=1}b_n\) diverges, the Limit Comparison Test gives no information. Similarly, if \(\frac{a_n}{b_n} \to \infty \) and \(\displaystyle \sum^ \infty _{n=1}b_n\) converges, the test also provides no information.

For example, consider the two series \(\displaystyle \sum_{n=1}^ \infty \frac{1}{\sqrt{n}}\) and \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^2}\). These series are both \(p\)-series with \(p=\frac{1}{2}\) and \(p=2\), respectively. Since \(p=\frac{1}{2} \lt 1\), the series \(\displaystyle \sum_{n=1}^ \infty \frac{1}{\sqrt{n}}\) diverges. On the other hand, since \(p=2 \gt 1\), the series \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^2}\) converges. However, suppose we attempted to apply the Limit Comparison Test, using the convergent \(p\)−series \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^3}\) as our comparison series. First, we see that

\[\dfrac{1/\sqrt{n}}{1/n^3}=\dfrac{n^3}{\sqrt{n}}=n^{5/2} \to \infty \; \text{ as } \;n \to \infty . \nonumber \]

Similarly, we see that

\[\dfrac{1/n^2}{1/n^3}=n \to \infty \; \text{ as } \;n \to \infty . \nonumber \]

Therefore, if \(\frac{a_n}{b_n} \to \infty \) when \(\displaystyle \sum_{n=1}^ \infty b_n\) converges, we do not gain any information on the convergence or divergence of \(\displaystyle \sum_{n=1}^ \infty a_n\).

Caution

Just as in the Series Comparison Test, the terms of the series used for the Limit Comparison Test must be positive.

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. \(\displaystyle \sum^ \infty _{n=1}\frac{1}{\sqrt{n}+1}\)
2. \(\displaystyle \sum^ \infty _{n=1}\frac{2^n+1}{3^n}\)
3. \(\displaystyle \sum^ \infty _{n=1}\frac{\ln(n)}{n^2}\)

Solution

a. Compare this series to \(\displaystyle \sum^ \infty _{n=1}\frac{1}{\sqrt{n}}\). Note that both series have only positive terms. Calculate

\[ \lim_{n \to \infty }\dfrac{1/(\sqrt{n}+1)}{1/\sqrt{n}}=\lim_{n \to \infty }\dfrac{\sqrt{n}}{\sqrt{n}+1}=\lim_{n \to \infty }\dfrac{1/\sqrt{n}}{1+1/\sqrt{n}}=1. \nonumber \]

By the Limit Comparison Test, since \(\displaystyle \sum^ \infty _{n=1}\frac{1}{\sqrt{n}}\) diverges, then \(\displaystyle \sum^ \infty _{n=1}\frac{1}{\sqrt{n}+1}\) diverges.

b. Compare this series to \(\displaystyle \sum^ \infty _{n=1}\left(\frac{2}{3}\right)^n\). Again, both series have only positive terms. We see that

\[ \lim_{n \to \infty }\dfrac{(2^n+1)/3^n}{2^n/3^n}=\lim_{n \to \infty }\dfrac{2^n+1}{3^n} \cdot \dfrac{3^n}{2^n}=\lim_{n \to \infty }\dfrac{2^n+1}{2^n}=\lim_{n \to \infty }\left[1+\left(\tfrac{1}{2}\right)^n\right]=1. \nonumber \]

Therefore,

\[ \lim_{n \to \infty }\dfrac{(2^n+1)/3^n}{2^n/3^n}=1. \nonumber \]

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

c. Since \(\ln n \lt n\) (see Example \( \PageIndex{1}c \)), compare with \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n}\). We see that

\[ \lim_{n \to \infty }\dfrac{\ln n/n^2}{1/n} = \lim_{n \to \infty }\dfrac{\ln n}{n^2} \cdot \dfrac{n}{1}=\lim_{n \to \infty }\dfrac{\ln n}{n}. \nonumber \]

In order to evaluate \(\displaystyle \lim_{n \to \infty }\ln n/n\), evaluate the limit as \(x \to \infty \) of the real-valued function \(\ln(x)/x\). These two limits are equal, and making this change allows us to use l’Hospital’s Rule.¹ We obtain

\[ \lim_{x \to \infty }\dfrac{\ln(x)}{x}=\lim_{x \to \infty }\dfrac{1}{x}=0. \nonumber \]

Therefore, \(\displaystyle \lim_{n \to \infty }\frac{\ln n}{n}=0\), and, consequently,

\[ \lim_{n \to \infty }\dfrac{(\ln n)/n^2}{1/n}=0. \nonumber \]

Since the limit is \(0\) but \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n}\) diverges, the Limit Comparison Test does not provide any information.

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

\[ \lim_{n \to \infty }\dfrac{(\ln n)/n^2}{1/n^2}=\lim_{n \to \infty }\dfrac{\ln n}{n^2} \cdot \dfrac{n^2}{1}=\lim_{n \to \infty }\ln n= \infty . \nonumber \]

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

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

\[ \lim_{n \to \infty }\dfrac{(\ln n)/n^2}{1/n^{3/2}}=\lim_{n \to \infty }\dfrac{\ln n}{n^2} \cdot \dfrac{n^{3/2}}{1}=\lim_{n \to \infty }\dfrac{\ln n}{\sqrt{n}}. \nonumber \]

As above, in order to evaluate \(\displaystyle \lim_{n \to \infty }\frac{\ln n}{\sqrt{n}}\), evaluate the limit as \(x \to \infty \) of the real-valued function \(\frac{\ln n}{\sqrt{n}}\). Using l’Hospital’s Rule,

\[ \lim_{x \to \infty }\dfrac{\ln x}{\sqrt{x}}=\lim_{x \to \infty }\dfrac{2\sqrt{x}}{x}=\lim_{x \to \infty }\dfrac{2}{\sqrt{x}}=0. \nonumber \]

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

Exercise \(\PageIndex{2}\)

Use the Limit Comparison Test to determine whether the series \(\displaystyle \sum^ \infty _{n=1}\frac{5^n}{3^n+2}\) converges or diverges.

Hint

Compare with a geometric series.

Answer

The series diverges.

Subsection Footnotes

¹ The move to considering the limit of the real-valued function \( \ln(x)/x \) is purposeful and showcases the need for rigor in Calculus. l'Hospital's Rule is only defined for continuous functions - not discontinuous functions like sequences. As such, we could not use l'Hospital's Rule on \( \displaystyle \lim_{n \to \infty} \ln(n)/n \).

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Key Concepts

• The Comparison Tests are used to determine convergence or divergence of series with positive terms.
• When using the Comparison Tests, a series \(\displaystyle \sum^ \infty _{n=1}a_n\) is often compared to a geometric or \(p\)-series.

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Glossary

Comparison Test

If \(0 \leq a_n \leq b_n\) for all \(n \geq N\) and \(\displaystyle \sum^ \infty _{n=1}b_n\) converges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) converges; if \(a_n \geq b_n \geq 0\) for all \(n \geq N\) and \(\displaystyle \sum^ \infty _{n=1}b_n\) diverges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) diverges.

Limit Comparison Test

Suppose \(a_n,b_n \geq 0\) for all \(n \geq 1\). If \(\displaystyle \lim_{n \to \infty }a_n/b_n \to L \neq 0\), then \(\displaystyle \sum^ \infty _{n=1}a_n\) and \(\displaystyle \sum^ \infty _{n=1}b_n\) both converge or both diverge; if \(\displaystyle \lim_{n \to \infty }a_n/b_n \to 0\) and \(\displaystyle \sum^ \infty _{n=1}b_n\) converges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) converges. If \(\displaystyle \lim_{n \to \infty }a_n/b_n \to \infty \), and \(\displaystyle \sum^ \infty _{n=1}b_n\) diverges, then \(\displaystyle \sum^ \infty _{n=1}a_n\) diverges.