Comparison Tests

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 11.4 – The Comparison Tests

Expected Educational Results

• Objective 20–01: I can analytically (i.e., not graphically and not numerically) verify all conditions needed to use a Comparison Test for a given series.

• Objective 20–02: I can use the Limit Comparison Test to determine if a series converges or diverges.

• Objective 20–03: I can use the Comparison Test to determine if a series converges or diverges.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Comparison Tests

Comparison Test

Definition: Comparison Test

Suppose $\sum a_n$$\sum{a_n}$ and $\sum b_n$$\sum{b_n}$ are series with positive terms.

• If $\sum b_n$$\sum{b_n}$ is convergent and $a_n\le b_n$$a_n\leq b_n$ for all $n$$n$, the $\sum a_n$$\sum{a_n}$ is also convergent.

• If $\sum b_n$$\sum{b_n}$ is divergent and $a_n\ge b_n$$a_n\geq b_n$ for all $n$$n$, the $\sum a_n$$\sum{a_n}$ is also divergent.

Explanation of Comparison Test

The Limit Comparison Test is written as a conditional statement.

1. “Suppose $\sum a_n$$\sum{a_n}$ and $\sum b_n$$\sum{b_n}$ are series with positive terms” is a requirement in order to use the Comparison Test.

   1. This requirement states all $a_n$$a_n$ must be positive and all $b_n$$b_n$ must be positive.

2. First conditional statement:

   1. “If $\sum b_n$$\sum{b_n}$ is convergent and $a_n\le b_n$$a_n\leq b_n$ for all $n$$n$​" is called the hypothesis of the conditional statement. You need to verify the hypothesis is true:

      1. $\sum b_n$$\sum{b_n}$ must be convergent, and;

      2. $a_n\le b_n$$a_n\leq b_n$ for all $n$$n$.

   2. “[then] the $\sum a_n$$\sum{a_n}$​ is also convergent” is called the conclusion of the conditional statement.

      1. The conclusion is True ONLY IF the hypothesis is True.

3. Second conditional statement:

   1. “If $\sum b_n$$\sum{b_n}$ is divergent and $a_n\ge b_n$$a_n\geq b_n$ for all $n$$n$​" is called the hypothesis of the conditional statement. You need to verify the hypothesis is true:

      1. $\sum b_n$$\sum{b_n}$ must be divergent, and;

      2. $a_n\ge b_n$$a_n\geq b_n$ for all $n$$n$.

   2. “[then] the $\sum a_n$$\sum{a_n}$​ is also divergent” is called the conclusion of the conditional statement.

      1. The conclusion is True ONLY IF the hypothesis is True.

NOTE: You must verify $a_n>0$$a_n>0$ and $b_n>0$$b_n>0$ before using the comparison tests. You must show work or an explanation on all Assessments.

NOTE: The Limit Comparison Test is typically much easier than the Comparison Test.

Example 02: Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n{e}^n}}$$\displaystyle \sum_{n=1}^{\infty}{a_n}=\sum_{n=1}^{\infty}{\frac{1}{ne^n}}$ converges or diverges using ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{e^n}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{e^n}}$

Solution:

Check the requirements:

$\sum a_n$$\sum{a_n}$: When $n\ge 1$$n\geq 1$, $e^n\ge 1\Rightarrow n{e}^n\ge 1\Rightarrow \frac{1}{n{e}^n}\ge 1>0$$e^n\geq 1\Rightarrow ne^n\geq 1\Rightarrow \frac{1}{ne^n}\geq 1>0$

$\sum b_n$$\sum{b_n}$: When $n\ge 1$$n\geq 1$, $e^n\ge 1\Rightarrow \frac{1}{e^n}\ge 1>0$$e^n\geq 1\Rightarrow \frac{1}{e^n}\geq 1>0$

So, the requirements are true.

Check the conditions:

1. If the comparison is being made to ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{e^n}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{e^n}}$, we know ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{e}\right)}^n}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\left(\frac{1}{e}\right)^n}$ which is a geometric series with $r=\frac{1}{e}$$r=\frac{1}{e}$. Since $-1<r<1$$-1<r<1$, $\sum b_n$$\sum{b_n}$ converges.

2. We need to show $a_n\le b_n$$a_n\leq b_n$:

${\displaystyle \Rightarrow \frac{1}{n{e}^n}\le \frac{1}{e^n}}$$\displaystyle \Rightarrow \frac{1}{ne^n}\leq\frac{1}{e^n}$

Multiply both sides of the inequality by $e^n$$e^n$ which is always positive so it does not change the inequality:

${\displaystyle \Rightarrow \cancel{e^n}\frac{1}{n\cancel{e^n}}\le \cancel{e^n}\frac{1}{\cancel{e^n}}\Rightarrow \frac{1}{n}\le 1\Rightarrow \cancel{n}\frac{1}{\cancel{n}}\le 1n\Rightarrow n\ge 1}$$\displaystyle \Rightarrow \cancel{e^n}\frac{1}{n\cancel{e^n}}\leq\cancel{e^n}\frac{1}{\cancel{e^n}}\Rightarrow \frac{1}{n}\leq 1\Rightarrow \cancel{n}\frac{1}{\cancel{n}}\leq 1\,n\Rightarrow n\geq 1$ which is a true statement from the index of the series.

Thus, $a_n\le b_n$$a_n\leq b_n$

Make the conclusion:

Both conditions of the Comparison Test are True, thus $\sum a_n$$\sum{a_n}$ is convergent by the Comparison Test.

Investigation 02

Use the given series, $\sum b_n$$\sum{b_n}$, to determine if the following series, $\sum a_n$$\sum{a_n}$, are convergent or divergent. Explain.

1. Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{{\sin }^2(n)}{n^2}}$$\displaystyle \sum_{n=1}^{\infty}{a_n}=\sum_{n=1}^{\infty}{\frac{\sin^2{(n)}}{n^2}}$ converges or diverges using ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{n^2}}$

2. Determine if ${\displaystyle \sum _{n=3}^{\mathrm{\infty }}{a}_n=\sum _{n=3}^{\mathrm{\infty }}\frac{\ln (n)}{n}}$$\displaystyle \sum_{n=3}^{\infty}{a_n}=\sum_{n=3}^{\infty}{\frac{\ln{(n)}}{n}}$ converges or diverges using ${\displaystyle \sum _{n=3}^{\mathrm{\infty }}{b}_n=\sum _{n=3}^{\mathrm{\infty }}\frac{1}{n}}$$\displaystyle \sum_{n=3}^{\infty}{b_n}=\sum_{n=3}^{\infty}{\frac{1}{n}}$

3. Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{2^{-1/n}}{n^3}}$$\displaystyle \sum_{n=1}^{\infty}{a_n}=\sum_{n=1}^{\infty}{\frac{2^{-1/n}}{n^3}}$ converges or diverges using ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^3}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{n^3}}$

4. Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{n^{-2}}{2^n}}$$\displaystyle \sum_{n=1}^{\infty}{a_n}=\sum_{n=1}^{\infty}{\frac{n^{-2}}{2^n}}$ converges or diverges using ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^n}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{2^n}}$

5. Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{n-1}{n^2\sqrt{n}+1}}$$\displaystyle \sum_{n=1}^{\infty}{a_n}=\sum_{n=1}^{\infty}{\frac{n-1}{n^2\sqrt{n}+1}}$ converges or diverges using ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{3/2}}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{n^{3/2}}}$

6. Determine if ${\displaystyle \sum _{n=4}^{\mathrm{\infty }}{a}_n=\sum _{n=4}^{\mathrm{\infty }}\frac{1}{n-3}}$$\displaystyle \sum_{n=4}^{\infty}{a_n}=\sum_{n=4}^{\infty}{\frac{1}{n-3}}$ converges or diverges using ${\displaystyle \sum _{n=4}^{\mathrm{\infty }}{b}_n=\sum _{n=4}^{\mathrm{\infty }}\frac{1}{n}}$$\displaystyle \sum_{n=4}^{\infty}{b_n}=\sum_{n=4}^{\infty}{\frac{1}{n}}$

7. Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4+e^n}}$$\displaystyle \sum_{n=1}^{\infty}{a_n}=\sum_{n=1}^{\infty}{\frac{1}{n^4+e^n}}$ converges or diverges using ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{n^4}}$

8. Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^4+e^n}}$$\displaystyle \sum_{n=1}^{\infty}{a_n}=\sum_{n=1}^{\infty}{\frac{1}{n^4+e^n}}$ converges or diverges using ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{e^n}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{e^n}}$

9. Determine if ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{2^n+1}{n{2}^n-1}}$$\displaystyle \sum_{n=1}^{\infty}{a_n}=\sum_{n=1}^{\infty}{\frac{2^n+1}{n2^n-1}}$ converges or diverges using ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{b}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}}$$\displaystyle \sum_{n=1}^{\infty}{b_n}=\sum_{n=1}^{\infty}{\frac{1}{n}}$

Use Technology to Verify Convergence or Divergence of a Series

NOTE: On all Assessments, you may use technology to verify your series convergence/divergence. You must show all Calculus II related work.

Mathematica

1
(* Example from Investigation 02-1 *)
2
SumConvergence[Sin[n]^2/n^2, n]

Warnings:

1. Be very careful with the syntax. Syntax is the set of rules on how to write computer code. Every software program has its own unique syntax. Some basic Mathematica syntax is located at: http://www.jjw3.com/TECH_Common_Functions.pdf.

2. SumConvergence[ ] has at least two arguments:

   1. The $n^{\text{th}}$$n^{\text{th}}$-term of the series;

   2. $n$$n$ where $n$$n$ is the index variable of the series.

3. For help on using the SumConvergence[ ] function:

   1. In Mathematica, execute the code: $\text{?SumConvergence}$$\text{?SumConvergence}$

   2. Click on $\vee$$\or$ near the bottom-left of output

   3. Click on local

   4. Read how to use the SumConvergence[ ] function – you will be able to copy-paste code.

4. Remember, correct Mathematica code will be all black except for variables.

5. To execute code (including comment codes), press and hold the SHIFT key and press the ENTER key.

6. Mathematica will return “True” (series converges) or “False” (series diverges).

Homework

At this time, you should be able to complete the following assignments:

• Section 11.4: # 1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Created: Tuesday, 17 November 2020 6:03 EDT Last Modified: Monday, 20 June 2022 - 12:16 (EDT)