You CAN Ace Calculus
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). |
external links you may find helpful |
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used to prove convergence |
yes |
used to prove divergence |
yes |
can be inconclusive |
no |
\( \sum{a_n} \) is the series we are trying to determine convergence or divergence, 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 \(n \geq N\). |
Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
Arc Length Parameter |
Conservative Vector Fields |
Cross Product |
Curl |
Curvature |
Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
Partial Derivatives |
Partial Integrals |
Path Integrals |
Potential Functions |
Principal Unit Normal Vector |
Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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Help Keep 17Calculus Free |
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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 | ||
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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 \( 0 < t_n \leq a_n \) |
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 |
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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 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 |
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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 |
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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: The Streetwise Guide, Including MultiVariable 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.
video by Dr Chris Tisdell
The first two minutes of this next video also contains a good explanation of the direct comparison test.
video by PatrickJMT
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 we explain 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.
video by Dr Chris Tisdell
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.
video by PatrickJMT
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new. |
Direct Comparison Test - Practice Problems Conversion |
[A01-160] - [A02-161] - [A03-163] - [A04-164] - [A05-165] - [A06-168] - [A07-169] - [A08-171] - [A09-222] |
[B01-157] - [B02-158] - [B03-159] - [B04-166] - [B05-170] - [B06-1868] - [B07-1869] |
[C01-162] - [C02-167] - [C03-172] |
Please update your notes to this new numbering system. The display of this conversion information is temporary. |
Instructions - - Unless otherwise instructed, determine the convergence or divergence of the following series using the direct comparison test, if possible.
Basic Problems |
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\(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{1}{k^4+3} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{1}{k^4+3} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \frac{1}{k^4+3} } }\)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{k=5}^{\infty}{ \frac{k^2}{k^3-1} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=5}^{\infty}{ \frac{k^2}{k^3-1} } }\)
Final Answer |
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The series diverges by the direct comparison test |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=5}^{\infty}{ \frac{k^2}{k^3-1} } }\)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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The series diverges by the direct comparison test |
close solution |
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2+3}}}\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2+3}}}\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2+3}}}\)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt{n}}{n^2+n} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt{n}}{n^2+n} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt{n}}{n^2+n} } }\)
Solution |
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There is a major mistake in this video. This video is included only to show you what NOT to do. In this video, she builds a table and compares values for small n and jumps to the conclusion that what she showed is true for ALL n. Although it worked with this problem, it won't work all the time and I've seen problems on exams where teachers expect you to use this technique and it doesn't work! So, how do we work this problem correctly?
The key comes in after she sets up the inequality. She wants to show that \(a_n \leq t_n \) for all n. Here is how to do this correctly.
\(\begin{array}{rcl}
a_n & \leq & t_n \\
\displaystyle{ \frac{\sqrt{n}}{n^2+n} } & \leq & \displaystyle{ \frac{1}{n^{3/2}} } \\
n^{1/2} n^{3/2} & \leq & n^2 + n \\
n^2 & \leq & n^2 + n \\
0 & \leq & n
\end{array}\)
Since \(n \geq 1\) (from the sum), then \(n \geq 0\) and so the last inequality above holds and therefore \(a_n \leq t_n \).
video by Krista King Math
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3+4} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3+4} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3+4} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^3+n+4} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^3+n+4} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^3+n+4} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{ \sqrt[3]{n^3+1}} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{ \sqrt[3]{n^3+1}} } }\)
Final Answer |
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The series diverges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{1}{ \sqrt[3]{n^3+1}} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series diverges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sin^2 n}{5^n} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sin^2 n}{5^n} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sin^2 n}{5^n} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\)
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\) diverges by the p-series test or the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\)
Solution |
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There are at least two ways to think about this one.
p-Series
First, if you factor out the constant in the numerator, you get \(\displaystyle{ 2 \sum_{n=1}^{\infty}{ \frac{1}{n} } }\)
and the remaining sum is just a p-series with \(p=1\). Therefore, the series diverges.
Direct Comparison Test
If you didn't pick up the fact that you have p-series, you can use the Direct Comparison Test, comparing the given series with
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }\)
which is a divergent p-series.
Now, to confirm that the given series diverges, we need to test \(\displaystyle{ \frac{2}{n} \geq \frac{1}{n} }\) for all n. If we multiply both sides of the inequality by n (which we can do since n is always positive), we get \( 2 \geq 1 \). This is true for all n, therefore the series diverges.
Final Answer |
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The series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n} } }\) diverges by the p-series test or the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{3/2}+1} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{3/2}+1} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{3/2}+1} } }\)
Solution |
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Final Answer |
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The series converges by the direct comparison test. |
close solution |
Intermediate Problems |
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\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{\ln(n)} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{\ln(n)} } }\)
Final Answer |
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The series diverges by the Direct Comparison Test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{\ln(n)} } }\)
Solution |
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This one is a little tricky. If we look at a graph of \( 1/ \ln(x) \) and compare it with the graph of \( 1/x \), it looks like \( 1/ \ln(x) \) is always greater than \( 1/x \). So our tendency is to use the Direct Comparison Test and just say that since \( 1/x \) diverges, so does \( 1/ \ln(x) \). This is certainly correct but how do you actually show that \( 1/ \ln(x) \) is always greater than \( 1/x \). I mean, isn't it possible that the graphs meet and maybe cross so that this isn't the case for some very large x? Let's see if we come up with an argument that holds for all x.
First, based on the logic above, we need to show that \(\displaystyle{ \frac{1}{\ln(x)} \geq \frac{1}{x} }\) holds.
Since \( x \) and \( \ln(x) \) are both positive, we can just cross multiply to get \( x \geq \ln(x) \)
Okay, let's call \( f(x)=x \) and \( g(x)=\ln(x) \).
We know that when \( x=1 \), \( f(1) = 1 \) and \( g(1) = 0\) . So the inequality holds for \( x=1 \). Now we will require that \( x > 1 \) (and so also \( n > 1 \)).
Now, \(f'(x) = 1\) and \(g'(x)=1/x\). Since the derivative is just the slope of the function, if we want \( f(x) \geq g(x) \) then we can just establish a point where the case holds (which we have done at \( x=1 \)) and as long as the slope of the smaller function is less than the slope of the larger function, then they should never cross and the smaller function will always be smaller than the larger function. Does this hold?
Well, can we show that \( f'(x) \geq g'(x) \)? Let's plug in the derivatives and see what happens. \( 1 \geq 1/x \to x \geq 1 \). This last inequality is something that we are already requiring. So the inequality holds and we have just shown that \( f(x) \geq g(x) \) for \( x \geq 1 \).
Therefore, by the Direct Comparison Test, since the test series \(\sum{1/x}\) diverges and the terms in the original series are always greater than (or equal to) the corresponding terms in the test series, the original series also diverges.
It will help you in working through this solution, if you have the graph in front of you. So take some extra time and graph \( y=1/x \) and \( y = 1/ \ln(x)\) on the same set of axes using your graphing utility.
video by Dr Chris Tisdell
Final Answer |
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The series diverges by the Direct Comparison Test. |
close solution |
\(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n\ln(n)-1} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n\ln(n)-1} } }\)
Final Answer |
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The series diverges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n\ln(n)-1} } }\)
Solution |
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We will use the direct comparison test and compare this series to \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{1}{n \ln( n )} } }\). 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 \( t_n < a_n \) for all \( n > 2 \).
\(\begin{array}{rcl}
\displaystyle{ \frac{1}{n \ln(n)} } & < & \displaystyle{ \frac{1}{n \ln(n)-1} } \\
n \ln(n)-1 & < & n \ln(n) \\
-1 & < & 0
\end{array}\)
Since the last inequality holds for all \(n\), then the first inequality holds and so \( t_n < a_n \) for all \( n > 2 \).
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.
Final Answer |
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The series diverges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(\cos(n))^3}{n^2+n+1} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(\cos(n))^3}{n^2+n+1} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{(\cos(n))^3}{n^2+n+1} } }\)
Solution |
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Step 1 - Choose A Test Series
First, let's look at the denominator polynomial \(n^2+n+1\). As \(n\) gets very large, the \(n^2\) term will dominate the other two terms. So we will drop \(n+1\) leaving only \(n^2\) in the denominator.
Now, let's look at the numerator. There is no easy way to determine what \((\cos(n))^3\) does as \(n\) goes to infinity. Essentially, it oscillates between \(-1\) and \(+1\). However, we know that \(\cos(n) \leq 1 \) and also \((\cos(n))^3 \leq 1\). Now, we need one more piece to this that may not be obvious. Looking at \(\cos(n)\), we need to make sure this will never be zero. (You will see why in a minute.) Can we say that? When IS \(\cos(x) = 0\)? This is zero when \(x\) is a multiple of \( \pi/2 \). In other words, does there exist a positive integer \(k\) such that \(n = k\pi/2\)? No, \(n = 1, 2, 3, . . .\) and \(\pi\) is irrational. This means that \(n\) will never be a multiple of \(\pi\) and, therefore, \(\cos(n)\) 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 \(a_n\) 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 \(\sum{\left|a_n\right|}\) converges, then \(\sum{a_n}\) also converges. So we will replace \(\cos(n)\) with \(\left|\cos(n)\right|\). So we will replace \((\cos(n))^3\) with \(1\) in our test series and see what happens.
So, this gives us the p-series \(\sum{1/n^2}\) as a test series which converges. Call this series \(\sum{t_n}\) where \(t_n = 1/n^2\)
Step 2 - Set up the inequality
Since the test series we ended up with converges, we set up the inequality as \(0 < a_n \leq t_n \).
Now I hope you see the importance of making sure that \(\cos(n)\) never evaluated to zero. If it did, the first part of the inequality \( 0 < a_n \) would not hold and so we would not be able to use this test.
So the inequality that we need to prove is
\(\displaystyle{ \frac{(\left|\cos(n)\right|)^3}{n^2+n+1} \leq \frac{1}{n^2} }\)
Step 3 - Prove the inequality holds
\(\displaystyle{ \frac{(\left|\cos(n)\right|)^3}{n^2+n+1} \leq \frac{1}{n^2} }\) |
\(\displaystyle{ (\left|\cos(n)\right|)^3 \leq \frac{n^2+n+1}{n^2} }\) |
\(\displaystyle{ (\left|\cos(n)\right|)^3 \leq 1 + 1/n + 1/n^2 }\) |
We discussed near the top of this solution that \( (\left|\cos(n)\right|)^3 \) 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 \(n\). Therefore, the series \(\sum{a_n}\) converges.
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{1+\sin(n)}{10^n} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{1+\sin(n)}{10^n} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{1+\sin(n)}{10^n} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{e^{3/n^2}}{n} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{e^{3/n^2}}{n} } }\)
Final Answer |
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The series diverges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{e^{3/n^2}}{n} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series diverges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln n}{n^3} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln n}{n^3} } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{\ln n}{n} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{\ln n}{n} } }\)
Solution |
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video by Dr Chris Tisdell
close solution |
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\sin(n)}{n^3+n+1}}}\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{\sin(n)}{n^3+n+1}}}\)
Final Answer |
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The series converges by the direct comparison test and it converges absolutely by the absolute convergence theorem. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{\sin(n)}{n^3+n+1}}}\)
Solution |
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\(\displaystyle{ a_n = \frac{\sin(n)}{n^3+n+1} }\)
Before we start with the direct comparison test, we have to look carefully at the requirements. Notice that in both inequalities, we require that \(a_n > 0 \). 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 \(\sum{\abs{a_n}}\) converges, then \(\sum{a_n}\) also converges. In this case,
\(\displaystyle{ \abs{a_n} = \abs{\frac{\sin(n)}{n^3+n+1}} = \frac{\abs{\sin(n)}}{\abs{n^3+n+1}} }\)
Since \(n > 0\) the denominator is always positive so we just need to replace \(\sin(n)\) with \(\abs{\sin(n)}\). 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 \(n\) the \(n^3\) term dominates. In the numerator, \(\abs{\sin(n)}\) is always less than or equal to one, let's compare this series with the test series \(\sum{t_n}\) where \(\displaystyle{t_n = \frac{1}{n^3} }\).
Since \(t_n\) is a p-series with \(p=3>1\), the test series converges. So the direct comparison test requires us to set up the inequality as \(a_n \leq t_n\).
\(\begin{array}{rcl}
a_n & \leq & t_n \\
\displaystyle{\frac{\abs{\sin(n)}}{n^3+n+1}} & \leq & \displaystyle{\frac{1}{n^3}} \\
\abs{\sin(n)} & \leq & \displaystyle{\frac{n^3+n+1}{n^3}} \\
\abs{\sin(n)} & \leq & \displaystyle{1 + \frac{1}{n^2} + \frac{1}{n^3}}
\end{array}\)
The last inequality is always true since \(\abs{\sin(n)}\leq 1\) and the right side is always greater than one. So the series converges.
The direct comparison test is the best test for this series. For a discussion on using other tests, see the practice problem page dedicated to this problem.
Final Answer |
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The series converges by the direct comparison test and it converges absolutely by the absolute convergence theorem. |
close solution |
\(\displaystyle{\sum_{n=0}^{\infty}{\frac{1}{3^n+n}}}\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{\sum_{n=0}^{\infty}{\frac{1}{3^n+n}}}\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{\sum_{n=0}^{\infty}{\frac{1}{3^n+n}}}\)
Solution |
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\(\displaystyle{ a_n = \frac{1}{3^n+n} }\)
Since \(3^n >> n\) for large \(n\), let's compare this series with the test series \(\sum{t_n}\) where \(\displaystyle{t_n = \frac{1}{3^n} }\).
Since \(t_n\) is a p-series with \(p=3>1\), the test series converges. So the direct comparison test requires us to set up the inequality as \(a_n < t_n\).
\(\begin{array}{rcl}
a_n & < & t_n \\
\displaystyle{\frac{1}{3^n+n}} & < & \displaystyle{\frac{1}{3^n}} \\
3^n & < & 3^n + n \\
0 & < & n
\end{array}\)
Conclusion
This last inequality is true for \(N=1\) and \(n>N\). So the series \(\sum{a_n}\) converges.
The direct comparison test is the best test for this series. For a discussion on using other tests, see the practice problem page dedicated to this problem.
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2-\cos^2(n)}}}\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2-\cos^2(n)}}}\)
Final Answer |
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The series diverges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2-\cos^2(n)}}}\)
Solution |
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\(\displaystyle{ a_n = \frac{n}{n^2-\cos^2(n)} }\)
For large \(n\) the \(n^2\) term dominates in the denominator since \(\cos^2(n) \leq 1 \). In the numerator, we have \(n\). So let's compare this series with the test series \(\sum{t_n}\) where \(\displaystyle{t_n = \frac{1}{n} }\).
Since \(t_n\) is a p-series with \(p=1\), the test series diverges. So the direct comparison test requires us to set up the inequality as \(t_n \leq a_n\).
\(\begin{array}{rcl}
t_n & \leq & a_n \\
\displaystyle{\frac{1}{n}} & \leq & \displaystyle{\frac{n}{n^2 - \cos^2(n)}} \\
n^2 - \cos^2(n) & \leq & n^2 \\
-\cos^2(n) & \leq & 0 \\
\cos^2(n) & \geq & 0
\end{array}\)
The last inequality is always true since a squared term is always greater than or equal to zero. So the series diverges.
The direct comparison test is the best test for this series. For a discussion on using other tests, see the practice problem page dedicated to this problem.
Final Answer |
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The series diverges by the direct comparison test. |
close solution |
Advanced Problems |
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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^2} } }\)
Solution |
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Although his logic and answer are correct in the video solution, he doesn't really show much work for this solution. So we will show it here.
We make the assumption that the series converges. This is reasonable since we have \(n^2\) in the denominator. So we need to choose a series that is larger than this series that also converges. We might be tempted to choose \(t_n = 1/n\) but this series diverges. So let's choose \(t_n = 1/n^{3/2}\). The series \(\sum{t_n}\) converges since it is a p-series with \(p=3/2>1\).
The direct comparison test requires that \(a_n \leq t_n\), which we need to show.
\(\begin{array}{rcl}
a_n & \leq & t_n \\
\displaystyle{ \frac{\ln n}{n^2} } & \leq & \displaystyle{ \frac{1}{n^{3/2}} } \\
\displaystyle{ \frac{\ln n}{n^{1/2}} } & \leq & 1
\end{array}\)
So how do we show that the last inequality holds. We will use the technique that calculates the derivative to show that it is always negative. Then, if we can find a value for n that is less than 1, then the inequality will hold.
First note that we need to take the derivative with respect to a continuous function. So we convert the variable to x. Taking the derivative with respect to n is not valid.
\(\displaystyle{ \frac{d}{dx}\left[ \frac{\ln x}{x^{1/2}} \right] }\) |
Use the quotient rule. |
\(\displaystyle{ \frac{x^{1/2}(1/x) - \ln x (1/2)x^{-1/2}}{x} }\) |
\(\displaystyle{ \frac{1-\ln x}{x^{3/2}} }\) |
Looking at the last fraction, we note that the denominator is always positive (since \(x>0\)) and for the same reason the numerator is alway negative. So the slope is negative. If we look at the point \(n=2\), \(\ln(2)/2^{1/2} \approx 0.490 \lt 1\). So we have shown that the inequality \(\displaystyle{ \frac{\ln n}{n^{1/2}} } \leq 1 \). Therefore the series converges by the direct comparison test.
Note: Using the limit comparison test may be easier.
video by MIT OCW
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n!}{n^n} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n!}{n^n} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n!}{n^n} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series converges by the direct comparison test. |
close solution |
\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\tan^{-1}(n)}{n^3} } }\)
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\tan^{-1}(n)}{n^3} } }\)
Final Answer |
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The series converges by the direct comparison test. |
Problem Statement |
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Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\tan^{-1}(n)}{n^3} } }\)
Solution |
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video by PatrickJMT
Final Answer |
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The series converges by the direct comparison test. |
close solution |