Calculus/Direct Comparison Test


Direct Comparison Test

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The Direct Comparison Test (DCT) 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.

Some instructors will tell you that this test is very basic and intuitive, but at first it can be difficult to understand and use for many students. However, once it is mastered, it is a quite powerful test and it will work more easily than the Limit Comparison Test since we don't need to evaluate a limit.

Key - - The key is to set up the inequality correctly. This can only be done once a test series is chosen and you know the convergence/divergence of that test series.

First we present the definition of the Direct Comparison Test and then we explain each part.

Direct Comparison Test Definition

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For the series   and the test series   where  

Convergence To prove convergence of  ,   must converge and  

Divergence To prove divergence of  ,   must diverge and  

Direct Comparison Test Quick Notes

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used to prove convergence yes
used to prove divergence yes
can be inconclusive no
  1. Notice that we do not specify the  -values on the sum. This is common in calculus and it just means that, for this test, it doesn't matter where the series starts (but it always 'ends' at infinity, since this is an infinite series).
  2.   is the series of which we are trying to determine convergence or divergence and it is given in the problem statement.
  3.   is the test series that you choose for comparison.
  4. In the course of using this test, you may need to find some real, finite value   where the inequalities hold for all  .
  5. Be careful when setting up the inequality. It is set up differently depending on whether you are assuming convergence or divergence.
  6. A subtle distinction for using this test that is implied in the inequalities in the above definition is that  . So be careful not to use this on a series that is alternating or that contains any negative terms.
  7. Some instructors may say that this test can be inconclusive. This is not correct. This test will always tell you if the series   either converges or diverges IF a valid test series can be found AND the inequality can be proven. There will be times when a valid test series cannot be found or when the inequality cannot be proven but this does not mean that the DCT is inconclusive.

How To Use the Direct Comparison Test

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There are three main steps to using this test.

Step 1 choose a test series
Step 2 set up the inequality
Step 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

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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   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   for   to replace   with   or use   for  .

As you get experience with this test, it will become easier to determine a good test series. So work plenty of practice problems.

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   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.

There is no general rule for choosing a test series but with some experience you will begin to see patterns and we will show you some examples and explain how to chose a test series (something that many/most books leave out).

Step 2 - Set Up The Inequality

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If you have a series   and you choose a test series   then you can set up the inequality in one of two ways:

Convergence Divergence
If you are assuming convergence, the test series must also converge and the inequality you need to prove is  . If you are assuming divergence, the test series must diverge and you need to prove is  .

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 as   increases 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 as   increases and not allowing it to go off to infinity.

Step 3 - Prove The Inequality Holds

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

Technique 1 - Directly

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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   by subtracting   from both sides to get  . This last inequality is always true.

NOTE: Be careful when squaring and taking square (even) roots. The resulting inequality may not be equal.

Technique 2 - Prove an inequality is always positive

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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   we can argue that, since   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   starts at zero or one and is always positive after that.

This technique also works if you can find a value   such that the expression holds for all  . Similar to the last example, you can use this argument for the inequality   by saying that for  , the inequality holds.

Technique 3 - Using Slope

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The third technique is to use the concept of slope and is best demonstrated with an example. Let's show that   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   is discrete (  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  , we can use   and for   we can use   where   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   for all   greater than some value. Let's call   and  . If we can show that   for some specific value of   and the slope of   is greater than the slope of  , then   will always be greater than  . The graphs will never cross and the inequality   will hold. You can see this intuitively in this graph (but you cannot use the graph to prove this).


 


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

Taking the derivatives, we have   and  . We need to show  

        

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

This shows that  .

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

What If The Inequality Doesn't Hold?

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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.

If the inequality doesn't hold, it doesn't mean that the Direct Comparison Test can't be used. We may just need to choose a different test series. One suggestion is to get a feel for what the series looks like is to plot the series on your calculator or use another test. Then see if you can extrapolate a smaller or larger series in order to come up with another test series.

How NOT To Use The Direct Comparison Test

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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   to show that the inequality holds for all  . 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 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 one video I saw, the presentor looked at the example   to show that you cannot use the direct comparison test by comparing this to   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  .

2. Use the integral test or, perhaps, the limit comparison test.

No matter what test you use, this series converges.

Study Tip

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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.

Be careful! Plotting and plugging in values is not a valid way to prove convergence or divergence of ANY series. Instructors anticipate this and will often put questions on exams that give incorrect answers this way.

Video Recommendations

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If you want a complete lecture on the Direct Comparison Test, this is a good video clip. Notice he calls this The Comparison Test, leaving out the word Direct. You need to watch only the first 40 minutes and 12 seconds. After that he discusses the Limit Comparison Test.

Here is a quick video explaining this test.

The first five and a half minutes of this video explains the direct comparison test very well.

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

This 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.

Application To Improper Integrals

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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.

Example Set 1 - Test For Divergence

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Use the divergent and monotonic harmonic series   to determine if   is divergent, if possible, for each of following series.

Example 1.1

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Since we are assuming divergence, the inequality we need to set up is   where   and  . Since  ,   is also greater than zero and therefore the left half of the inequality holds. so we just need to show the right half of the inequality.

  

This last inequality holds for all  . Therefore the series diverges by the DCT.

This is a great example of how to choose a test series. Constants become negligible as   becomes very large. So sometimes choosing a test series by dropping constants work. Of course, this will not work all the time, which the next example shows.

Example 1.2

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As in the previous example, we only need to show that the right half of the inequality holds.

       

However, this is never true so this test series cannot be used. (Some instructors might say this means that the DCT is inconclusive but actually, this is not a valid test series to prove divergence.) The test series with   would be a better test for comparison.

Example 1.3

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As in the previous two examples, we only need to show that the right half of the inequality holds.

       

The last inequality holds for all  . So the series diverges by the DCT.

Example 1.4

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As in the previous examples, we only need to show that the right half of the inequality holds.

       

The last inequality holds for all  . So the series diverges by the DCT.

Two comments are in order here.

  1. Notice that we needed to find a value   where the inequality held for all values  .
  2. Although in general it is not a good idea to square both sides of an inequality and assume it is a valid operation, it works here since   is always greater than one.

Example 1.5

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As in the previous examples, we only need to show that the right half of the inequality holds.

 

The last inequality is not true for all   greater than some  , i.e. once   gets larger than 1, the last inequality no longer holds. Therefore, this test series cannot be used to prove divergence using the DCT.

Note that the DCT can still be used on this series but since the series converges, another test series is required if we want to use the DCT.

Example Set 2 - Test for Convergence

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Use the convergent and monotonic series   to determine if   is convergent, if possible, for each of following series.

Example 2.1

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  decreases at a faster rate than  . However, these series do not satisfy the   requirement, because   is larger than   when  . We can solve this issue by taking removing the first term from the both series to obtain   and  . Now, comparing   with   shows that   is indeed convergent. Because this is convergent, adding the original   will not change whether it is convergent or not, it will add   to the value of convergence.

Example 2.2

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  is smaller   for every  , so this series is convergent.

Example 2.3

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  is less than or equal to   and is greater than   for every   in the domain; this is because   conforms to </math>\frac{1}{2^n}</math>, and the fact that   is squared implies that it will never be less than zero.

Example 2.4

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  is convergent. Notice that this is just  , which is just   multiplied by some finite number.

Example 2.5

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  is greater than   for an infinite amount of  . So this test series cannot be used to test for convergence.

Example 2.6

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This series has negative terms, so the DCT cannot be used here.

Direct Comparison Test Practice Problems

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Practice Problems with Written Solutions

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Determine the convergence or divergence of these series using the Direct Comparison Test, if possible. If the DCT is inconclusive, use another test to determine convergence or divergence. Make sure to specify what test you used in your final answer.

1.  

Hint
Choose  
Choose  
Answer
The series diverges by the DCT.
The series diverges by the DCT.
Solution
We will use the direct comparison test and compare this series to  . This is a very common way to choose a series for comparison. Since constants, regardless of their values, become negligible when compared to very, very large numbers, we can often drop them and use the result as a test series. This won't always work and, of course, it depends on whether the test series converges or diverges, but it is a good place to start.
The test series is shown to diverge as a practice problem on the integral test page. So we just need to show that   for all  .
 
Since the last inequality holds for all  , then the first inequality holds and so   for all  .
Therefore, by the direct comparison test, since the test series diverges and its terms are less than the terms in the original series, the original series also diverges.
We will use the direct comparison test and compare this series to  . This is a very common way to choose a series for comparison. Since constants, regardless of their values, become negligible when compared to very, very large numbers, we can often drop them and use the result as a test series. This won't always work and, of course, it depends on whether the test series converges or diverges, but it is a good place to start.
The test series is shown to diverge as a practice problem on the integral test page. So we just need to show that   for all  .
 
Since the last inequality holds for all  , then the first inequality holds and so   for all  .
Therefore, by the direct comparison test, since the test series diverges and its terms are less than the terms in the original series, the original series also diverges.


2.  

Hint
Choose   and use the absolute convergence theorem to replace   with   in order to make the terms positive.
Choose   and use the absolute convergence theorem to replace   with   in order to make the terms positive.
Answer
The series converges by the DCT.
The series converges by the DCT.
Solution
Step 1 - Choose A Test Series

First, let's look at the denominator polynomial  . As   gets very large, the   term will dominate the other two terms. So we will drop   leaving only   in the denominator. Now, let's look at the numerator. There is no easy way to determine what   does as   goes to infinity. Essentially, it oscillates between   and  .

However, we know that   and also  . Now, we need one more piece to this that may not be obvious. Looking at  , we need to make sure this will never be zero. (You will see why in a minute.) Can we say that? When IS  ? This is zero when   is a multiple of  . In other words, does there exist a positive integer   such that  ? No,   and   is irrational. This means that   will never be a multiple of   and, therefore,   will never be zero.

However, there is one additional detail that we need to deal with before we can use this test. Notice that both inequalities require that the   terms be positive. This is an important detail. If we don't, we can't use the direct comparison test.

Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if   converges, then   also converges. So we will replace   with  . So we will replace   with   in our test series and see what happens.

So, this gives us the p-series   as a test series which converges. Call this series   where  

Step 2 - Set up the inequality

Since the test series we ended up with converges, we set up the inequality as  . Now I hope you see the importance of making sure that   never evaluated to zero. If it did, the first part of the inequality   would not hold and so we would not be able to use this test.

So the inequality that we need to prove is  

Step 3 - Prove the inequality holds

 

We discussed near the top of this solution that   is always less than one. Since the right side of the inequality is always greater than one, we can say that this inequality holds for all  . Therefore, the series   converges.
Step 1 - Choose A Test Series

First, let's look at the denominator polynomial  . As   gets very large, the   term will dominate the other two terms. So we will drop   leaving only   in the denominator. Now, let's look at the numerator. There is no easy way to determine what   does as   goes to infinity. Essentially, it oscillates between   and  .

However, we know that   and also  . Now, we need one more piece to this that may not be obvious. Looking at  , we need to make sure this will never be zero. (You will see why in a minute.) Can we say that? When IS  ? This is zero when   is a multiple of  . In other words, does there exist a positive integer   such that  ? No,   and   is irrational. This means that   will never be a multiple of   and, therefore,   will never be zero.

However, there is one additional detail that we need to deal with before we can use this test. Notice that both inequalities require that the   terms be positive. This is an important detail. If we don't, we can't use the direct comparison test.

Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if   converges, then   also converges. So we will replace   with  . So we will replace   with   in our test series and see what happens.

So, this gives us the p-series   as a test series which converges. Call this series   where  

Step 2 - Set up the inequality

Since the test series we ended up with converges, we set up the inequality as  . Now I hope you see the importance of making sure that   never evaluated to zero. If it did, the first part of the inequality   would not hold and so we would not be able to use this test.

So the inequality that we need to prove is  

Step 3 - Prove the inequality holds

 

We discussed near the top of this solution that   is always less than one. Since the right side of the inequality is always greater than one, we can say that this inequality holds for all  . Therefore, the series   converges.


3.  


Hint
Choose   and use the absolute convergence theorem to replace   with  .
Choose   and use the absolute convergence theorem to replace   with  .
Answer
The series converges by the DCT.
The series converges by the DCT.
Solution
 

Before we start with the direct comparison test, we have to look carefully at the requirements. Notice that in both inequalities, we require that  . However, for this series, the sine term introduces some negative terms. So we can't just directly use the direct comparison test. This is an important detail. If we don't handle this detail, we can't use the direct comparison test.

Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if   converges, then   also converges. In this case,
 
Since   the denominator is always positive so we just need to replace   with  . So, in this problem, we will work toward convergence. If we get the result that the series diverges, then we can't use the absolute convergence theorem and the result is inconclusive.

For large   the   term dominates. In the numerator,   is always less than or equal to one, let's compare this series with the test series   where  .

Since   is a p-series with  , the test series converges. So the direct comparison test requires us to set up the inequality as  .

 

The last inequality is always true since   and the right side is always greater than one. So the series converges.

 

Before we start with the direct comparison test, we have to look carefully at the requirements. Notice that in both inequalities, we require that  . However, for this series, the sine term introduces some negative terms. So we can't just directly use the direct comparison test. This is an important detail. If we don't handle this detail, we can't use the direct comparison test.

Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if   converges, then   also converges. In this case,
 
Since   the denominator is always positive so we just need to replace   with  . So, in this problem, we will work toward convergence. If we get the result that the series diverges, then we can't use the absolute convergence theorem and the result is inconclusive.

For large   the   term dominates. In the numerator,   is always less than or equal to one, let's compare this series with the test series   where  .

Since   is a p-series with  , the test series converges. So the direct comparison test requires us to set up the inequality as  .

 

The last inequality is always true since   and the right side is always greater than one. So the series converges.

Practice Problems with Video Solutions

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1  
answer
converges
converges

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2  
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converges
converges
solution