You CAN Ace Calculus

 infinite limits infinite series basics p-series geometric series Note: Although it is not absolutely necessary to know the Direct 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).

### Limit Comparison Test Quick Breakdown

 used to prove convergence yes used to prove divergence yes can be inconclusive yes $$\sum{a_n}$$ is the series we are trying to determine convergence or divergence $$\sum{t_n}$$ is the test series

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 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
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17calculus > infinite series > limit comparison test

 Limit Comparison Test What The Limit Comparison Test Says When To Use The Limit Comparison Test How To Choose A Test Series Practice

The Limit Comparison Test and the Direct Comparison Test are two tests where you choose a series that you know about and compare it to the series you are working with, that you don't know about, to determine convergence or divergence. These two tests are the next most important, after the Ratio Test, and it will help you to know these well. They are very powerful and fairly easy to use.

Limit Comparison Test

For the two series $$\sum{a_n}$$ and $$\sum{t_n}$$ where $$a_n > 0$$ and $$t_n > 0$$, we calculate $$\displaystyle{\lim_{n \to \infty}{\left[ \frac{a_n}{t_n} \right]} = L}$$
1. when L is finite and positive, the two series either both converge or both diverge
2. when $$L=0$$ and $$\sum{t_n}$$ converges, then $$\sum{a_n}$$ also converges
3. when $$L=\infty$$ and $$\sum{t_n}$$ diverges, then $$\sum{a_n}$$ also diverges
4. if the limit does not exist then the test is inconclusive

 What The Limit Comparison Test Says

This test is pretty straightforward. In our notation, we say that the series that you are trying to determine whether it converges or diverges is $$\sum{a_n}$$ and the test series that you know whether it converges or diverges is $$\sum{t_n}$$. The limit $$L$$ has be defined for you come to any conclusion.

Also, notice that the fraction can be inverted and the test still works for case 1 (but not cases 2 and 3). For example, if you get $$L=3$$ for one fraction, then you would get $$L=1/3$$ for the other fraction. Both are finite and positive and both will tell you whether your series converges or diverges. If you invert the fraction then cases 2 and 3 will change. So it is important to check your fraction if you are trying to apply cases 2 or 3.

 When To Use The Limit Comparison Test

This test can't be used all the time. Here is what to check before trying this test.
- As with all theorems, the conditions for the test must be met. The main one is that $$a_n > 0$$. If this isn't the case with your series, don't stop yet. Look at the Absolute Convergence Theorem to see if that will help you.
- A second important thing to consider is whether or not the limit can be evaluated. For example, if you have a term that oscillates, like sine or cosine, then you can't evaluate the limit. So this test won't help you. In this case, the Direct Comparison Test might work better.

 How To 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.

Before getting started working practice problems, watching these two video clips will give you a better feel for the limit comparison test. Both clips are short and to the point.

### Dr Chris Tisdell - Series, comparison + ratio tests [2min-19secs]

video by Dr Chris Tisdell

### PatrickJMT - Direct Comparison Test / Limit Comparison Test for Series - Basic Info [1min-35secs]

video by PatrickJMT

### Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

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

Basic Problems

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }$$

The series diverges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }$$

Solution

There are several ways to determine convergence or divergence of this infinite series. First, we notice that as n gets very, very large, the 1 in the denominator becomes negligible when compared to $$n^2$$. So the fraction is getting closer and closer to $$n/n^2 = 1/n$$. So let's try comparing the original series with $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n} } }$$.
I find the Limit Comparison Test to be the easiest to use. So let's set up the limit.

 $$\displaystyle{ \lim_{n \to \infty}{ \left( \frac{n}{n^2+1} \div \frac{1}{n} \right) } }$$ $$\displaystyle{ \lim_{n \to \infty}{ \left( \frac{n}{n^2+1} \cdot \frac{n}{1} \right) } }$$ $$\displaystyle{ \lim_{n \to \infty}{\frac{n^2}{n^2+1}} = 1 }$$

Since the limit is positive and finite, the two series either both converge or both diverge. The series $$\sum{1/n}$$ diverges since it is a p-series with $$p=1$$, which means the original series also diverges by the Limit Comparison Test.
Note - The Direct Comparison Test and Integral Test also work.

The series diverges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{n^2+1} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{n^2+1} } }$$

The series converges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{n^2+1} } }$$

Solution

There are several ways to determine convergence or divergence of this infinite series. First, we notice that as n gets very, very large, the 1 in the denominator becomes negligible when compared to $$n^2$$. So the fraction is getting closer and closer to $$1/n^2$$. So let's try comparing the original series with $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2} } }$$.
I find the Limit Comparison Test to be the easiest to use. So let's set up the limit.

 $$\displaystyle{ \lim_{n \to \infty}{ \left( \frac{3}{n^2+1} \div \frac{1}{n^2} \right) } }$$ $$\displaystyle{ \lim_{n \to \infty}{ \left( \frac{3}{n^2+1} \cdot \frac{n^2}{1} \right) } }$$ $$\displaystyle{ \lim_{n \to \infty}{\frac{3n^2}{n^2+1}} = 3 }$$

Since the limit is positive and finite, the two series either both converge or both diverge. The series $$\sum{1/n^2}$$ converges since it is a p-series with $$p=2$$, which means the original series also converges by the Limit Comparison Test.
The Direct Comparison Test and Integral Test may also work.

The series converges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{5n+2}{n^3+1} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{5n+2}{n^3+1} } }$$

The series converges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=0}^{\infty}{ \frac{5n+2}{n^3+1} } }$$

Solution

### 135 solution video

video by MIT OCW

The series converges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n^2-4}} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n^2-4}} } }$$

The series diverges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n^2-4}} } }$$

Solution

### 136 solution video

video by Dr Chris Tisdell

The series diverges by the Limit Comparison Test

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

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3n+1}{(2n-1)^4} } }$$

The series converges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3n+1}{(2n-1)^4} } }$$

Solution

### 138 solution video

video by Dr Chris Tisdell

The series converges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{6n-1}} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{6n-1}} } }$$

The series diverges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{6n-1}} } }$$

Solution

### 139 solution video

video by Dr Chris Tisdell

The series diverges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{10n^2}{n^3-1} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{10n^2}{n^3-1} } }$$

The series diverges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{10n^2}{n^3-1} } }$$

Solution

### 140 solution video

video by Krista King Math

The series diverges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{4^n-3} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{4^n-3} } }$$

The series converges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{4^n-3} } }$$

Solution

### 141 solution video

video by MIP4U

The series converges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{\sqrt{n^2-1}} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{\sqrt{n^2-1}} } }$$

The series diverges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{\sqrt{n^2-1}} } }$$

Solution

### 142 solution video

video by MIP4U

The series diverges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^4+2n^2-1}{6n^6+4n^4} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^4+2n^2-1}{6n^6+4n^4} } }$$

The series converges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^4+2n^2-1}{6n^6+4n^4} } }$$

Solution

### 143 solution video

video by MIP4U

The series converges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{2/3}+1} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{2/3}+1} } }$$

The series diverges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{2/3}+1} } }$$

Solution

### 144 solution video

video by Dr Phil Clark

The series diverges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3-4} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3-4} } }$$

The series converges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3-4} } }$$

Solution

### 146 solution video

video by PatrickJMT

The series converges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{\sqrt{n}}{n-1} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{\sqrt{n}}{n-1} } }$$

The series diverges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=2}^{\infty}{ \frac{\sqrt{n}}{n-1} } }$$

Solution

### 147 solution video

video by PatrickJMT

The series diverges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{2n^2+3} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{2n^2+3} } }$$

The series converges by the Limit Comparison Test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{2n^2+3} } }$$

Solution

### 149 solution video

video by PatrickJMT

The series converges by the Limit Comparison Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3+1}{n^4-n+5} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3+1}{n^4-n+5} } }$$

The series diverges by the Limit Comparison Test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3+1}{n^4-n+5} } }$$

Solution

### 150 solution video

video by PatrickJMT

The series diverges by the Limit Comparison Test.

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

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{n+1}{3n^3(4n-1)}}}$$

The series converges by the Limit Comparison Test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{n+1}{3n^3(4n-1)}}}$$

Solution

### 152 solution video

video by PatrickJMT

The series converges by the Limit Comparison Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3-n+14}{n^5+n^3+1} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3-n+14}{n^5+n^3+1} } }$$

The series converges by the Limit Comparison Test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3-n+14}{n^5+n^3+1} } }$$

Solution

### 154 solution video

video by PatrickJMT

The series converges by the Limit Comparison Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+4} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+4} } }$$

The series converges by the limit comparison test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+4} } }$$

Solution

Compare to $$\sum{1/n^2}$$, which is a convergent p-series with $$p = 2$$
$$\displaystyle{a_n=\frac{1}{n^2+4}}$$, $$\displaystyle{t_n=\frac{1}{n^2}}$$

$$\begin{array}{rcl} \displaystyle{ \lim_{n\to\infty}{ \frac{a_n}{t_n} } } & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{1}{n^2+4} \cdot \frac{n^2}{1} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^2}{n^2+4} \frac{1/n^2}{1/n^2} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \frac{1}{1+4/n^2} } = 1 } \end{array}$$

Since $$\displaystyle{ 0 \leq \lim_{n\to\infty}{ \frac{a_n}{t_n} } = 1 \lt \infty }$$ and $$\sum{t_n}$$ converges, the series also converges.
The limit comparison test is one of the best tests for this series. For a discussion on using other tests, see the practice problem page dedicated to this problem.

The series converges by the limit comparison test.

$$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2-1}{n^3+4}}}$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2-1}{n^3+4}}}$$

The series diverges by the limit comparison test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2-1}{n^3+4}}}$$

Solution

Use the test series $$\sum{1/n}$$, a divergent p-series.
$$\begin{array}{rcl} \displaystyle{ \lim_{n\to\infty}{ \left| \frac{a_n}{t_n} \right| } } & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^2-1}{n^3+4} \frac{n}{1} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^3-n}{n^3+4} \frac{1/n^3}{1/n^3} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{1-1/n^2}{1+4/n^3} \right] } = 1} \\ \end{array}$$
Since the limit is finite and positive, both series either converge or diverge. Since the test series diverges, so does the original series.

The series diverges by the limit comparison test.

Intermediate Problems

$$\displaystyle{\sum_{n=1}^{\infty}{\frac{2\sqrt{n+3}}{n+5}}}$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{2\sqrt{n+3}}{n+5}}}$$

The series diverges by the Limit Comparison Test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{\sum_{n=1}^{\infty}{\frac{2\sqrt{n+3}}{n+5}}}$$

Solution

### 151 solution video

video by PatrickJMT

The series diverges by the Limit Comparison Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4n^2}{\sqrt{9n^4+1}} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4n^2}{\sqrt{9n^4+1}} } }$$

The series diverges by the Divergence Test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4n^2}{\sqrt{9n^4+1}} } }$$

Solution

### 153 solution video

video by PatrickJMT

The series diverges by the Divergence Test.

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

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^{3/2}} } }$$

The series converges by the Limit Comparison Test

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^{3/2}} } }$$

Solution

### 137 solution video

video by Dr Chris Tisdell

The series converges by the Limit Comparison Test

$$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt[3]{n}+\ln(n^4)}{n} } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt[3]{n}+\ln(n^4)}{n} } }$$

The series diverges by the Limit Comparison Test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt[3]{n}+\ln(n^4)}{n} } }$$

Solution

### 155 solution video

video by PatrickJMT

The series diverges by the Limit Comparison Test.

$$\displaystyle{ \sum_{n=1}^{\infty}{ \sin(1/n) } }$$

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \sin(1/n) } }$$

The series diverges by the Limit Comparison Test.

Problem Statement

Determine convergence or divergence of the series $$\displaystyle{ \sum_{n=1}^{\infty}{ \sin(1/n) } }$$

Solution

### 156 solution video

video by PatrickJMT