Comparison Tests

Calculus 2

Overview

Earlier in the course, we saw that a given improper integral converges if its integrand is less than the integrand of another integral known to converge. Similarly, a given improper integral diverges if its integrand is greater than the integrand of another integral known to diverge. We apply a similar strategy to determine whether certain series converge or diverge.

An ideal series to use a comparison test on is a series that "looks like" a series whose convergence is known -- a series with the same dominant term(s). We will often compare the given series to a geometric series or $p$ -series. Note that there may be more than one appropriate series to compare to.

Direct Comparison Test (DCT)

An analogy that works great when explaining the Direct Comparison Test (DCT) is a pet door:

If I know a cat easily fits through the door, then I know that any cat of equal or smaller size would also fit through the door. However, I do not know for sure whether a cat of larger size would fit.
If I know a dog does not fit through the door, then I know that any dog of equal or larger size would also not fit through the door. However, I do not know for sure whether a smaller dog would fit.

We compare series that are "smaller" or "larger" than series with known convergence to see if they fit through this mathematical pet door.

Therefore, the Direct Comparison Test (DCT) states that if $0 \leq a_n \leq b_n$ for all $n \geq N$ , where $N \in \mathbb{N}$ , then:

If $\sum_{n=1}^\infty b_n$ converges, then so does $\sum_{n=1}^\infty a_n$ . (That is, if the bigger series converges, the smaller series also converges)
If $\sum_{n=1}^\infty a_n$ diverges, then so does $\sum_{n=1}^\infty b_n$ . (That is, if the smaller series diverges, the bigger series also diverges)

Keep in mind that no other claims can be made with the DCT:

If the bigger series diverges, we know nothing about the smaller series.
If the smaller series converges, we know nothing about the bigger series.

Example 1

Let $a_n = \frac{1}{2^n+n}$ and $b_n = \left(\frac{1}{2}\right)^n$ .

Does $\sum_{n=1}^\infty b_n$ converge or diverge? Why?
How do the sizes of the terms $a_n$ and $b_n$ compare?
What can be concluded about $\sum_{n=1}^\infty \frac{1}{2^n+n}$ ?

View Answer

$\sum_{n=1}^\infty b_n$ converges because it is a geometric series with $r = \frac{1}{2} < 1$ .
$a_n \leq b_n$ for all $n$ .
By the DCT, $\sum_{n=1}^\infty a_n$ also converges.

Example 2

Let $a_n = \frac{1}{n^2+n+1}$ .

By considering the rate of growth of the denominator of $a_n$ , what choice would you make for $b_n$ ?
Does $\sum_{n=1}^\infty b_n$ converge or diverge? Why?
How do the sizes of the terms $a_n$ and $b_n$ compare?
What can be concluded about $\sum_{n=1}^\infty \frac{1}{n^2+n+1}$ ?

View Answer

Choose $b_n = \frac{1}{n^2}$ because the dominant term in the denominator of $a_n$ is $n^2$ .
$\sum_{n=1}^\infty b_n$ converges by the p-Series Test ( $p = 2 > 1$ ).
$a_n \leq b_n$ for all $n$ .
By the DCT, $\sum_{n=1}^\infty a_n$ also converges.

Example 3

Use the Direct Comparison Test to determine whether $\sum_{n=1}^\infty \frac{\sqrt{n^4-1}}{n^5+3}$ converges or diverges.

View Answer

Compare with $b_n = \frac{n^2}{n^5} = \frac{1}{n^3}$ . Since $p = 3 > 1$ , $\sum_{n=1}^\infty b_n$ converges, and $a_n \leq b_n$ , the series converges by the DCT.

Example 4

Use the Direct Comparison Test to determine whether $\sum_{k=1}^\infty \frac{1}{3^k+1}$ converges or diverges.

View Answer

Compare with $b_k = \frac{1}{3^k}$ . Since $b_k$ is a geometric series with $r = \frac{1}{3} < 1$ , $\sum_{k=1}^\infty b_k$ converges, and $a_k \leq b_k$ , the series converges by the DCT.

Example 5

Use the Direct Comparison Test to determine whether $\sum_{n=2}^\infty \frac{5}{n-1}$ converges or diverges.

View Answer

Compare with $b_n = \frac{5}{n}$ . Since $b_n$ is the harmonic series (scaled by 5), which diverges, and $a_n \geq b_n$ , the series diverges by the DCT.

Example 6

Use the Direct Comparison Test to determine whether $\sum_{n=1}^\infty \frac{n}{n^4-1}$ converges or diverges.

View Answer

Compare with $b_n = \frac{n}{n^4} = \frac{1}{n^3}$ . Since $p = 3 > 1$ , $\sum_{n=1}^\infty b_n$ converges, and $a_n \leq b_n$ , the series converges by the DCT.

Limit Comparison Test (LCT)

The last example illustrated that the Direct Comparison Test might not always help us conclude convergence or divergence (whether that is because the inequality goes the wrong way or the inequality itself is difficult to prove).

If we strongly suspect that the terms of the series will behave like another series with known convergence, we can use the Limit Comparison Test.

The Limit Comparison Test states that if $a_n > 0$ and $b_n > 0$ for all $n$ , and if $\lim_{n \to \infty} \frac{a_n}{b_n} = L,$ where $L$ is finite and $L > 0$ , then the two series $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

Example 7

Use the Limit Comparison Test to determine whether $\sum_{n=1}^\infty \frac{n}{n^4-1}$ converges or diverges.

View Answer

Converges

View Solution

Compare with $b_n = \frac{1}{n^3}$ . Since:

\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n}{n^4-1} \cdot n^3 = \lim_{n \to \infty} \frac{n^4}{n^4-1} = 1,

and $\sum b_n$ converges by the p-Series Test, $\sum a_n$ also converges.

Example 8

Use either the Direct or Limit Comparison Test to determine whether $\sum_{n=2}^\infty \frac{n^3-2n}{n^4+3}$ converges or diverges.

View Answer

Diverges

View Solution

Compare with $b_n = \frac{1}{n}$ . Since:

\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n^3-2n}{n^4+3} \cdot n = \lim_{n \to \infty} \frac{n^4-2n^2}{n^4+3} = 1,

and $\sum b_n$ diverges (harmonic series), $\sum a_n$ also diverges.

Example 9

Use either the Direct or Limit Comparison Test to determine whether $\sum_{k=1}^\infty \frac{4^k}{5^k-6}$ converges or diverges.

View Answer

Converges

View Solution

Compare with $b_k = \left(\frac{4}{5}\right)^k$ . Since:

\lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{4^k}{5^k-6} \cdot \frac{5^k}{4^k} = \lim_{k \to \infty} \frac{5^k}{5^k-6} = 1,

and $\sum b_k$ converges (geometric series with $r = \frac{4}{5} < 1$ ), $\sum a_k$ also converges.

Example 10

Determine whether $\sum_{n=1}^\infty \frac{10n+1}{n(n+1)(n+2)}$ converges or diverges.

View Answer

Converges

Example 11

Determine whether $\sum_{n=1}^\infty \frac{2\sqrt{n}}{n^{3/2}+1}$ converges or diverges.

View Answer

Diverges

p-Series

Integral Test