9 Sequences and Series 9.3 The Integral Test 9.5 Alternating Series and Absolute Convergence

9.4 Comparison Tests

In this section we will be comparing a given series with series that we know either converge or diverge.

Theorem 9.4.1 Direct Comparison Test

Let $\{a_{n}\}$ and $\{b_{n}\}$ be positive sequences where $a_{n}\leq b_{n}$ for all $n\geq N$ , for some $N\geq 1$ .

1.

If $\displaystyle\sum_{n=1}^{\infty}b_{n}$ converges, then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges.
2.

If $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges, then $\displaystyle\sum_{n=1}^{\infty}b_{n}$ diverges.

^†^†margin: Note: A sequence

\{a_{n}\}

is a positive sequence if

a_{n}>0

for all

n

. Because of Theorem 9.2.5, any theorem that relies on a positive sequence still holds true when

a_{n}>0

for all but a finite number of values of

n

Proof

First consider the partial sums of each series.

$S_{n}=\sum_{i=1}^{n}a_{i}\quad\text{and}\quad T_{n}=\sum_{i=1}^{n}b_{i}$

Since both series have positive terms we know that

$S_{n}\leq S_{n}+a_{n+1}=\sum_{i=1}^{n}a_{i}+a_{n+1}=\sum_{i=1}^{n+1}a_{i}=S_{n% +1}$

and

$T_{n}\leq T_{n}+b_{n+1}=\sum_{i=1}^{n}b_{i}+b_{n+1}=\sum_{i=1}^{n+1}b_{i}=T_{n% +1}$

Therefore, both of the sequences of partial sums, $\{S_{n}\}$ and $\{T_{n}\}$ , are increasing.

For $n\geq N$ , We’re now going to split each series into two parts:

$\displaystyle S$ $\displaystyle=\sum_{i=1}^{N-1}a_{i}$ $\displaystyle T$ $\displaystyle=\sum_{i=1}^{N-1}b_{i}$

$\displaystyle\bar{S}_{n}$ $\displaystyle=\sum_{i=N}^{n}a_{i}$ $\displaystyle\bar{T}_{n}$ $\displaystyle=\sum_{i=N}^{n}b_{i}.$

This means that $S_{n}=S+\bar{S}_{n}$ and $T_{n}=T+\bar{T}_{n}$ . Also, because $a_{n}\leq b_{n}$ for all $n\geq N$ , we must have $\bar{S}_{n}\leq\bar{T}_{n}$ .

For the first part of the theorem, assume that $\displaystyle\sum_{n=1}^{\infty}b_{n}$ converges. Since $b_{n}\geq 0$ we know that

$T_{n}=\sum_{i=1}^{n}b_{i}\leq\sum_{i=1}^{\infty}b_{i}$

From above we know that $\bar{S}_{n}\leq\bar{T}_{n}$ for all $n\geq N$ so we also have

$S_{n}=S+\bar{S}_{n}\leq S+\bar{T}_{n}=S+T_{n}-T=S-T+\sum_{i=1}^{\infty}b_{i}$

Because $\displaystyle\sum_{i=1}^{\infty}b_{i}$ converges it must have a finite value and $\{S_{n}\}$ is bounded above. We also showed that $\{S_{n}\}$ is increasing so by Theorem 9.1.6 we know $\{S_{n}\}$ converges and so $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges.

For the second part, assume that $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges. Because $a_{n}\geq 0$ we must have $\displaystyle\lim_{n\to\infty}S_{n}=\infty$ . We also know that for all $n\geq N$ , $\bar{S}_{n}\leq\bar{T}_{n}$ . This means that

$\displaystyle\lim_{n\to\infty}T_{n}=\lim_{n\to\infty}(T+\bar{T}_{n})\\ \displaystyle\geq T+\lim_{n\to\infty}\bar{S}_{n}=T+\lim_{n\to\infty}(S_{n}-S)=% T-S+\lim_{n\to\infty}S_{n}=\infty.$

Therefore, $\{T_{n}\}$ is a divergent sequence and so $\displaystyle\sum_{i=1}^{\infty}b_{n}$ diverges. ∎

Watch the video:
Direct Comparison Test / Limit Comparison Test for Series — Basic Info from https://youtu.be/LAHKu3B-1zE

Example 9.4.1 Applying the Direct Comparison Test

Determine the convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}+n^{2}}$ .

SolutionThis 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 $3^{n}<3^{n}+n^{2}$ , $\displaystyle\frac{1}{3^{n}}>\frac{1}{3^{n}+n^{2}}$ for all $n\geq 1$ . The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}}$ is a convergent geometric series; by Theorem 9.4.1, $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}+n^{2}}$ converges.

Example 9.4.2 Applying the Direct Comparison Test

Determine the convergence of $\displaystyle\sum_{n=2}^{\infty}\frac{n^{3}}{n^{4}-1}$ .

SolutionWe know the Harmonic Series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ diverges, and it seems that the given series is closely related to it, hence we predict it will diverge.

We have $\displaystyle\frac{n^{3}}{n^{4}-1}>\frac{n^{3}}{n^{4}}=\frac{1}{n}$ for all $n\geq 1$ .

The Harmonic Series, $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ , diverges, so we conclude that $\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}}{n^{4}-1}$ diverges as well.

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 $\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}}{n^{4}+1}$ . It is very similar to the divergent series given in Example 9.4.2. We suspect that it also diverges, as $\displaystyle\frac{1}{n}\approx\frac{n^{3}}{n^{4}+1}$ for large $n$ . However, the inequality that we naturally want to use “goes the wrong way”: since $\displaystyle\frac{n^{3}}{n^{4}+1}<\frac{n^{3}}{n^{4}}=\frac{1}{n}$ for all $n\geq 1$ . 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.

Limit Comparison Test

Theorem 9.4.2 Limit Comparison Test

Let $\{a_{n}\}$ and $\{b_{n}\}$ be positive sequences.

1.

If $\displaystyle\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L$ , where $L$ is a positive real number, then $\displaystyle\sum_{n=1}^{\infty}a_{n}$ and $\displaystyle\sum_{n=1}^{\infty}b_{n}$ either both converge or both diverge.
2.

If $\displaystyle\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=0$ , then if $\displaystyle\sum_{n=1}^{\infty}b_{n}$ converges, then so does $\displaystyle\sum_{n=1}^{\infty}a_{n}$ .
3.

If $\displaystyle\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\infty$ , then if $\displaystyle\sum_{n=1}^{\infty}b_{n}$ diverges, then so does $\displaystyle\sum_{n=1}^{\infty}a_{n}$ .

Proof
1. 1.
  
  We have $0<L<\infty$ so we can find two positive numbers, $m$ and $M$ such that $m<L<M$ . Because $\displaystyle L=\lim_{n\to\infty}\frac{a_{n}}{b_{n}}$ we know that for large enough $n$ the quotient $\frac{a_{n}}{b_{n}}$ must be close to $L$ . So there must be a positive integer $N$ such that if $n>N$ we also have $\displaystyle m<\frac{a_{n}}{b_{n}}<M$ . Multiply by $b_{n}$ and we have $mb_{n}<a_{n}<Mb_{n}$ for $n>N$ . If $\displaystyle\sum_{n=1}^{\infty}b_{n}$ diverges, then so does $\displaystyle\sum_{n=1}^{\infty}mb_{n}$ . Also since $mb_{n}<a_{n}$ for sufficiently large $n$ , by the Comparison Test $\displaystyle\sum_{n=1}^{\infty}a_{n}$ also diverges.
  Similarly, if $\displaystyle\sum_{n=1}^{\infty}b_{n}$ converges, then so does $\displaystyle\sum_{n=1}^{\infty}Mb_{n}$ . Since $a_{n}<Mb_{n}$ for sufficiently large $n$ , by the Comparison Test $\displaystyle\sum_{n=1}^{\infty}a_{n}$ also converges.
2. 2.
  
  Since $\displaystyle\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=0$ , there is a number $N>0$ such that
  
  $\displaystyle\left\lvert\frac{a_{n}}{b_{n}}-0\right\rvert$ $\displaystyle<1\text{ for all }n>N$
  
  $\displaystyle a_{n}$ $\displaystyle<b_{n}\text{ since }a_{n}\text{ and }b_{n}\text{ are positive}$
  
  Now since $\displaystyle\sum_{n=1}^{\infty}b_{n}$ converges, $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges by the Comparison Test.
3. 3.
  
  Since $\displaystyle\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\infty$ , there is a number $N>0$ such that
  
  $\displaystyle\frac{a_{n}}{b_{n}}$ $\displaystyle>1\text{ for all }n>N$
  
  $\displaystyle a_{n}$ $\displaystyle>b_{n}\text{ for all }n>N$
  
  Now since $\displaystyle\sum_{n=1}^{\infty}b_{n}$ diverges, $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges by the Comparison Test.∎

Theorem 9.4.2 is most useful when the convergence of the series from $\{b_{n}\}$ is known and we are trying to determine the convergence of the series from $\{a_{n}\}$ .

We use the Limit Comparison Test in the next example to examine the series $\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}}{n^{4}+1}$ which motivated this new test.

Example 9.4.3 Applying the Limit Comparison Test

Determine the convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}}{n^{4}+1}$ using the Limit Comparison Test.

SolutionWe compare the terms of $\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}}{n^{4}+1}$ to the terms of the Harmonic Sequence $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ :

	$\displaystyle\lim_{n\to\infty}\frac{{n^{3}}/(n^{4}+1)}{1/n}$	$\displaystyle=\lim_{n\to\infty}\frac{n^{4}}{n^{4}+1}=\lim_{n\to\infty}\frac{1}% {1+1/n^{4}}$
		$\displaystyle=1.$

Since the Harmonic Series diverges, we conclude that $\displaystyle\sum_{n=1}^{\infty}\frac{n^{3}}{n^{4}+1}$ diverges as well.

Example 9.4.4 Applying the Limit Comparison Test

Determine the convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}-n^{2}}$

SolutionThis series is similar to the one in Example 9.4.1, but now we are considering “ $3^{n}-n^{2}$ ” instead of “ $3^{n}+n^{2}$ .” This difference makes applying the Direct Comparison Test difficult.

Instead, we use the Limit Comparison Test and compare with the series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}}$ :

	$\displaystyle\lim_{n\to\infty}\frac{1/(3^{n}-n^{2})}{1/3^{n}}$	$\displaystyle=\lim_{n\to\infty}\frac{3^{n}}{3^{n}-n^{2}}$
		$\displaystyle\mathrel{\overset{\text{by LHR}}{=}}\lim_{n\to\infty}\frac{\ln 3% \cdot 3^{n}}{\ln 3\cdot 3^{n}-2n}$
		$\displaystyle\mathrel{\overset{\text{by LHR}}{=}}\lim_{n\to\infty}\frac{(\ln 3% )^{2}3^{n}}{(\ln 3)^{2}3^{n}-2}$
		$\displaystyle\mathrel{\overset{\text{by LHR}}{=}}\lim_{n\to\infty}\frac{(\ln 3% )^{3}3^{n}}{(\ln 3)^{3}3^{n}}=1.$

We know $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}}$ is a convergent geometric series, hence $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}-n^{2}}$ converges as well.

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 $\{a_{n}\}$ . It is also helpful to note that factorials dominate increasing exponentials, which dominate algebraic functions (e.g., polynomials), which dominate logarithms. In the previous example, the dominant term of $\displaystyle\frac{1}{3^{n}-n^{2}}$ was $3^{n}$ , so we compared the series to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}}$ . It is hard to apply the Limit Comparison Test to series containing factorials, though, as we have not learned how to apply L’Hôpital’s Rule to $n!$ .

Example 9.4.5 Applying the Limit Comparison Test

Determine the convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt{n}+3}{n^{2}-n+1}$ .

SolutionWe naïvely attempt to apply the rule of thumb given above and note that the dominant term in the expression of the series is $1/n^{2}$ . Knowing that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ converges, we attempt to apply the Limit Comparison Test:

	$\displaystyle\lim_{n\to\infty}\frac{(\sqrt{n}+3)/(n^{2}-n+1)}{1/n^{2}}$	$\displaystyle=\lim_{n\to\infty}\frac{n^{2}(\sqrt{n}+3)}{n^{2}-n+1}$
		$\displaystyle=\infty\quad\text{(Apply L'H\^{o}pital's Rule)}.$

Theorem 9.4.2 part (3) only applies when $\displaystyle\sum_{n=1}^{\infty}b_{n}$ 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 $n^{1/2}$ and the dominant term of the denominator is $n^{2}$ . Thus we should compare the terms of the given series to $n^{1/2}/n^{2}=1/n^{3/2}$ :

	$\displaystyle\lim_{n\to\infty}\frac{(\sqrt{n}+3)/(n^{2}-n+1)}{1/n^{3/2}}$	$\displaystyle=\lim_{n\to\infty}\frac{n^{3/2}(\sqrt{n}+3)}{n^{2}-n+1}$
		$\displaystyle=1\quad\text{(Apply L'H\^{o}pital's Rule)}.$

Since the $p$ -series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ converges, we conclude that $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt{n}+3}{n^{2}-n+1}$ converges as well.

The tests we have encountered so far have required that we analyze series from positive sequences (the absolute value of the ratio and the root tests of Section 9.6 will convert the sequence into a positive sequence). The next section relaxes this restriction by considering alternating series, where the underlying sequence has terms that alternate between being positive and negative.

Exercises 9.4

Terms and Concepts

1.

Suppose $\displaystyle\sum_{n=0}^{\infty}a_{n}$ is convergent, and there are sequences $\{b_{n}\}$ and $\{c_{n}\}$ such that $0\leq b_{n}\leq a_{n}\leq c_{n}$ for all $n$ . What can be said about the series $\displaystyle\sum_{n=0}^{\infty}b_{n}$ and $\displaystyle\sum_{n=0}^{\infty}c_{n}$ ?
2.

Suppose $\displaystyle\sum_{n=0}^{\infty}a_{n}$ is divergent, and there are sequences $\{b_{n}\}$ and $\{c_{n}\}$ such that $0\leq b_{n}\leq a_{n}\leq c_{n}$ for all $n$ . What can be said about the series $\displaystyle\sum_{n=0}^{\infty}b_{n}$ and $\displaystyle\sum_{n=0}^{\infty}c_{n}$ ?

Problems

In Exercises 3–8, use the Direct Comparison Test to determine the convergence of the given series; state what series is used for comparison.

3.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}+3n-5}$
4.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{4^{n}+n^{2}-n}$
5.

$\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n}$
6.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n!+n}$
7.

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{\sqrt{n^{2}-1}}$
8.

$\displaystyle\sum_{n=5}^{\infty}\frac{1}{\sqrt{n}-2}$

In Exercises 9–14, use the Limit Comparison Test to determine the convergence of the given series; state what series is used for comparison.

9.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}-3n+5}$
10.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{4^{n}-n^{2}}$
11.

$\displaystyle\sum_{n=4}^{\infty}\frac{\ln n}{n-3}$
12.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^{2}+n}}$
13.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n+\sqrt{n}}$
14.

$\displaystyle\sum_{n=1}^{\infty}\sin\big{(}1/n\big{)}$

In Exercises 15–24, use the Direct Comparison Test or the Limit Comparison Test to determine the convergence of the given series. State which series is used for comparison.

15.

$\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}+n+1}{n^{3}-5}$
16.

$\displaystyle\sum_{n=1}^{\infty}\frac{n-10}{n^{2}+10n+10}$
17.

$\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}}{5^{n}+10}$
18.

$\displaystyle\sum_{n=1}^{\infty}\frac{n+5}{n^{3}-5}$
19.

$\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^{4}+1}$
20.

$\displaystyle\sum_{n=1}^{\infty}\frac{n-1}{n4^{n}}$
21.

$\displaystyle\sum_{n=2}^{\infty}\frac{n}{n^{2}-1}$
22.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}+100}$
23.

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n^{2}\ln n}$
24.

$\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt{n}+3}{n^{2}+17}$

In Exercises 25–32, determine the convergence of the given series. State the test used; more than one test may be appropriate.

25.

$\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}}{2^{n}}$
26.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n+5)^{3}}$
27.

$\displaystyle\sum_{n=1}^{\infty}\frac{n!}{10^{n}}$
28.

$\displaystyle\sum_{n=1}^{\infty}\frac{\ln n}{n!}$
29.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n}+n}$
30.

$\displaystyle\sum_{n=1}^{\infty}\frac{n-2}{10n+5}$
31.

$\displaystyle\sum_{n=1}^{\infty}\frac{3^{n}}{n^{3}}$
32.

$\displaystyle\sum_{n=1}^{\infty}\frac{\cos(1/n)}{\sqrt{n}}$
33.
Given that $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges, state which of the following series converges, may converge, or does not converge. (a) $\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}}{n}$ (b) $\displaystyle\sum_{n=1}^{\infty}a_{n}a_{n+1}$ (c) $\displaystyle\sum_{n=1}^{\infty}(a_{n})^{2}$ (d) $\displaystyle\sum_{n=1}^{\infty}na_{n}$ (e) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{a_{n}}$

	$\displaystyle S$	$\displaystyle=\sum_{i=1}^{N-1}a_{i}$	$\displaystyle T$	$\displaystyle=\sum_{i=1}^{N-1}b_{i}$
	$\displaystyle\bar{S}_{n}$	$\displaystyle=\sum_{i=N}^{n}a_{i}$	$\displaystyle\bar{T}_{n}$	$\displaystyle=\sum_{i=N}^{n}b_{i}.$

	$\displaystyle\left\lvert\frac{a_{n}}{b_{n}}-0\right\rvert$	$\displaystyle<1\text{ for all }n>N$
	$\displaystyle a_{n}$	$\displaystyle<b_{n}\text{ since }a_{n}\text{ and }b_{n}\text{ are positive}$

	$\displaystyle\frac{a_{n}}{b_{n}}$	$\displaystyle>1\text{ for all }n>N$
	$\displaystyle a_{n}$	$\displaystyle>b_{n}\text{ for all }n>N$