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You CAN Ace Calculus

17calculus > infinite series > direct comparison test

Topics You Need To Understand For This Page

infinite series basics

p-series

geometric series

Note: Although it is not absolutely necessary to know the Limit Comparison Test for this page, the two comparison tests are closely related. Many times ( but not always ), both techniques will work when determining the convergence or divergence of a series. So which technique you choose is often personal preference (assuming you are given the opportunity to choose).

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Related Topics and Links

Direct Comparison Test

on this page: ► how to use the test     ► how NOT to use the test     ► study tip     ► application to improper integrals

The Direct Comparison Test is sometimes simply called The Comparison Test. However, like we do here, many books include the word 'Direct' in the name to clearly separate this test from the Limit Comparison Test.

Direct Comparison Test

For the series \( \sum{a_n} \) and test series \( \sum{t_n} \) where \( t_n > 0 \)

to prove convergence of \( \sum{a_n} \)

 

to prove divergence of \( \sum{a_n} \)

\( \sum{t_n} \) must converge and \( 0 < a_n \leq t_n \)

 

\( \sum{t_n} \) must diverge and \( 0 < t_n \leq a_n \)

Quick Summary and Notes

used to prove convergence

yes

used to prove divergence

yes

can be inconclusive

no

In the notation above, \( \sum{a_n} \) is the series we are trying to determine the convergence or divergence of, and \( \sum{t_n} \) is the test series.

In the course of using this test, we may need to find some real, finite value \(N>0\) where the inequalities hold for all \(n \geq N\).

Key - - This test can be difficult to understand and use for many students. The key is to set up the inequality correctly. This can only be done once a test series is chosen and you make an assumption about the convergence/divergence based on that test series.

How To Use The Direct Comparison Test

There are three main steps to using this test.
1. choose a test series
2. set up the inequality
3. prove the inequality holds

The difficult thing about this test is that it seems like you are expected to already know whether the series converges or diverges before you even use the test. It helps to have a feel for it (and, with enough practice, you will develop this over time) but if you don't, you can guess. If you reach a dead-end when trying to prove the inequality holds, try to prove the other direction. You will find more suggestions on this in step 1, below.
Let's look at the details of each step.

Step 1 - Choose A Test Series

When you are first learning this technique, it may look like the test series comes out of thin air and you just randomly choose one and see if it works. If it doesn't, you try another one. This is not the best way to choose a test series. The best way I've found is to use the series you are asked to work with and come up with the test series. There are several things to consider.

The first key is to choose a test series that you know converges or diverges AND that will help you get a finite, positive limit.
Idea 1: If you have polynomials in both the numerator and denominator of a fraction, drop all terms except for the highest power terms (in both parts) and simplify. Drop any constants. What you end up with may be a good comparison series. The reason this works is that, as n gets larger and larger, the highest powers dominate. You will often end up with a p-series that you know either converges or diverges.
Idea 2: Choose a p-series or geometric series since you can tell right away whether it converges or diverges.
Idea 3: If you have a sine or cosine term, you are always guaranteed that the result is less than or equal to one and greater than or equal to negative one. If you don't have any bounds on the angle, these are the best you can do. So replace the sine or cosine term with one.
Idea 4: If you have a natural log, use the fact that \(\ln(n) \leq n\) for \(n\geq1\) to replace \(\ln(n)\) with n or use \(\ln(n) \geq 1\) for \(n\geq3\) .

As you get experience with this test, it will become easier to determine a good test series. So work plenty of practice problems and follow the advice in the infinite series study techniques section.

The additional thing you need to think about for the Direct Comparison Test that doesn't matter for the Limit Comparison Test is that you need to have a feel for whether the series \(\sum{a_n}\) converges or diverges. This requires a certain amount of experience, since it determines how you set up the inequality. But you can guess by looking at the test series you ended up with. If it diverges, then your series may too. Similarly, if the test series converges, then you want to test for convergence of your original series.

Step 2 - Set Up The Inequality

If you have a series \( \sum{a_n} \) and you choose a test series \( \sum{t_n} \) then you can set up the inequality in one of two ways:

If you are assuming convergence, the test series must also converge and the inequality you need to show is \( 0 < a_n \leq t_n \).

If you are assuming divergence, the test series must diverge and you need to show \( 0 < t_n \leq a_n \).

Here is an idea on how to think about the direction of the inequality. If you think the series you are working with diverges, you want to pick a divergent test series that is SMALLER than the series you are working with. You can think about this smaller test series as 'pushing up' your series and since the small series diverges, there is no way your series can converge since it is always being pushed up to infinity.

However, if you think your series converges, then you need to choose a convergent test series that is LARGER than your series. Then, you can think about the test series as always 'holding down' your series and not allowing it to go off to infinity.

Step 3 - Prove The Inequality Holds

Once you get the inequality set up, you need to prove that the inequality holds for all n greater than some N. There are several techniques to do this depending on the inequality, one of which should work.

Technique 1 - Directly
First, directly. In this case, set up the inequality and perform algebraic operations until you get an inequality that always holds. For example, we can show that \( n \leq n+2 \) by subtracting n from both sides to get \( 0 \leq 2 \). This last inequality is always true.

Technique 2 - Prove an inequality is always positive.
If the direct way is not possible, try moving all the terms to one side leaving zero on the other side and then explain how the expression is always positive. For example, if we can get something like \( 0 \leq \displaystyle{ \frac{n+5}{n^3} }\) we can argue that, since n is always positive, both the numerator and denominator are positive, resulting in the right side always being greater than zero. The key to remember here is that n starts at zero or one and is always positive after that.

This technique also works if you can find a value N such that the expression holds for all \( n \geq N \). Similar to the last example, you can use this argument for the inequality \( 0 \leq \displaystyle{ \frac{n-5}{n^3} }\)by saying that for \( n \geq 6 \), the inequality holds.

Technique 3 - Using Slope
The third technique is to use the concept of slope and is best demonstrated with an example. Let's show that \( n \geq \ln(n) \) using slope.

The first thing to remember about slope is, to find the slope, you take the derivative and the derivative is defined only on continuous functions. In our case \(n\) is discrete (\(n\) takes on the discrete values 1, 2, 3, 4, ... but no value in between these numbers.), so we need to find continuous functions that have the same values at the discrete values. We don't care what is going on between the discrete values as long as the function is continuous. So for \(n\), we can use \(f(x)=x\) and for \( \ln(n) \) we can use \( g(x)=\ln(x) \) where \(x\) is a continuous variable in both functions. Now we have continuous functions so that we can take derivatives. I know this is a fine point, but we need to get it right.

Okay, we need to show that \( x \geq \ln(x) \)for all \(x\) greater than some value. Let's call \( f(x) = x \)and \( g(x) = \ln(x) \). If we can show that \( f(x) \geq g(x) \) for some specific value of \(x\) and the slope of \( f(x) \) is greater than the slope of \( g(x) \), then \( f(x) \) will always be greater than \( g(x) \). The graphs will never cross and the inequality \( x \geq \ln(x) \) will hold. You can see this intuitively in the graph on the right.

Let's see if we can show this. First, we know that when \( x = 1\), \( f(1) = 1 \)and \( g(1) = 0 \). Since \( 1 \geq 0 \), we have established a point (\(x=1\)) where \( f(x) \geq g(x) \). Now we will use slope to establish that the functions stay that way for all \(x > 1 \).

Taking the derivatives, we have \( f'(x) = 1\)and \( g'(x) = 1/x \).
We need to show \( f'(x) \geq g'(x) \)

\(\displaystyle{ \begin{array}{rcl} f'(x) & \geq & g'(x) \\ 1 & \geq & 1/x \\ x & \geq & 1 \end{array} }\)

Now, since, \(x\) is always greater than or equal to \(1\), then the slope of \(f(x)\) is always larger than the slope of \(g(x)\). This says that \(f(x)\) is increasing at a faster rate than \(g(x)\) and therefore will always be larger.

To recap, what we have done here is that we have found a point, \(x=1\) where the inequality holds. Then we have used slope to say that the inequality holds for all values greater than that value.

If you are unable to prove the inequality, then you either need to choose a different test series or try another test. Using the Direct Comparison Test takes practice and time to sink in before you can understand it and use it. Another good description of how to think about this can be found in the book How To Ace The Rest Of Calculus.

Okay, now that you have read the above information, you have a general idea what the direct comparison test is and how to use it. Now let's watch a couple of videos to make the ideas clearer. Watching both of these is important to help you understand and use this test.

The first five and a half minutes of this first video explains the direct comparison test very well. This is one of the best instructors ever.

Dr Chris Tisdell: Series, comparison + ratio tests (part 1)

The first two minutes of this next video also contains a good explanation of the direct comparison test.

PatrickJMT: Direct Comparison Test / Limit Comparison Test for Series - Basic Info

How NOT To Use The Direct Comparison Test

There are two main ways students might try to use the Direct Comparison Test that do not work.
1. Build a table using the first few values of n to show that the inequality holds for all n. The video associated with one of the practice problems shows someone doing this and explains how to correctly show that the inequality holds. Be very, very careful to not use this technique. It is a pitfall that instructors watch for.
2. Set up the inequality incorrectly. The rest of this section, including the video below, shows this and explains why it doesn't work.

Here is how NOT to use the direct comparison test, i.e. when this test does not work when the inequality is set up incorrectly. In the video below, he looks at the example \(\displaystyle{ \sum_{n=2}^{\infty}{\frac{1}{n(\ln(n))^2}} }\) to show that you cannot use the direct comparison test by comparing this to \(\displaystyle{ \sum{\frac{1}{n^3}} }\) to prove convergence. When this happens, you have two choices.
1. You can choose a different test series.
2. You can try another test.

In this example, either of these choices will work.
1. Compare this series to \(\displaystyle{ \sum{\frac{1}{n^2}} }\).
2. Use the integral test or, perhaps, the limit comparison test.

No matter what test you use, this series converges.

This next video clip is important to watch since it shows a pitfall that almost every student falls into when using this test and most teachers will watch for.

Dr Chris Tisdell: Series, comparison + ratio tests (part 2)

Study Tip

When learning this test, it may help you to draw graphs of what is going on. Although this is a good technique to use in general, it will especially help you with this test, since the inequalities can best be shown in a graph. You do not even need specific functions. You can just draw generic functions that are above and below a test function.

Application To Improper Integrals

Even if you have not had improper integrals yet, this video is excellent to watch anyway to help you visualize the direct comparison test. You don't need to understand improper integrals to get a lot out of this video.

PatrickJMT: Direct Comparison Test for ( Improper ) Integrals

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Practice Problems

Instructions - - Unless otherwise instructed, determine the convergence or divergence of the following series using the direct comparison test, if possible.

Level A - Basic

Practice A01

\(\displaystyle{\sum_{k=1}^{\infty}{\frac{1}{k^4+3}}}\)

answer

solution

Practice A02

\(\displaystyle{\sum_{k=5}^{\infty}{\frac{k^2}{k^3-1}}}\)

answer

solution

Practice A03

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

answer

solution

Practice A04

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\sqrt{n}}{n^2+n}}}\)

answer

solution

Practice A05

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{2}{n^3+4}}}\)

answer

solution

Practice A06

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^3+n+4}}}\)

answer

solution

Practice A07

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{\sqrt[3]{n^3+1}}}}\)

answer

solution

Practice A08

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\sin^2 n}{5^n}}}\)

answer

solution

Practice A09

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{2}{n}}}\)

answer

solution


Level B - Intermediate

Practice B01

\(\displaystyle{\sum_{n=2}^{\infty}{\frac{1}{\ln(n)}}}\)

answer

solution

Practice B02

\(\displaystyle{\sum_{n=2}^{\infty}{\frac{1}{n\ln(n)-1}}}\)

answer

solution

Practice B03

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{(\cos(n))^3}{n^2+n+1}}}\)

answer

solution

Practice B04

\(\displaystyle{\sum_{n=0}^{\infty}{\frac{1+\sin(n)}{10^n}}}\)

answer

solution

Practice B05

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{e^{3/n^2}}{n}}}\)

answer

solution

Practice B06

\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{\ln n}{n^3} } }\)

solution

Practice B07

\(\displaystyle{ \sum_{n=3}^{\infty}{\frac{\ln n}{n} } }\)

solution


Level C - Advanced

Practice C01

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\ln(n)}{n^2}}}\)

answer

solution

Practice C02

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n!}{n^n}}}\)

answer

solution

Practice C03

\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\tan^{-1}(n)}{n^3}}}\)

answer

solution

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