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17Calculus Infinite Series - Limit Comparison Test

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

The Limit Comparison Test (LCT) 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 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.

If you would like a complete lecture on the Limit Comparison Test, we recommend this video clip. As the title implies, this video starts with the (Direct) Comparison Test but we have the video start when he begins discussing the Limit Comparison Test.

Prof Leonard - Calculus 2 Lecture 9.4: The Comparison Test for Series and The Limit Comparison Test

video by Prof Leonard

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

Limit Comparison Test Proof

Here is a video showing a proof of the Limit Comparison Test. You do not need to watch this in order to understand and use the Limit Comparison Test. However, we include it here for those of you who are interested.

Linda Green - Proof of the Limit Comparison Test

video by Linda Green

How to Read and Do Proofs: An Introduction to Mathematical Thought Processes

Practice

Unless otherwise instructed, determine the convergence or divergence of the following series using the limit comparison test (LCT), if possible. [Of course, if the LCT is inconclusive, use another test to determine convergence or divergence.]

Basic

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n}{n^2+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

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.

Ludus - 133 video solution

video by Ludus

Final Answer

The series diverges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{n^2+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{n^2+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

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.

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{5n+2}{n^3+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{5n+2}{n^3+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

MIT OCW - 135 video solution

video by MIT OCW

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n+3}{n(n+2)} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Ludus - 3214 video solution

video by Ludus

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n+1}{2n^3-n^2} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

We skip over the first minute and a half or so of this video to where he actually starts solving this problem. However, in that first minute and a half, he discusses why the Direct Comparison Test does not work to solve this problem. If you have had the Direct Comparison Test, you may want to go back and watch that first part of the video. However, you do not need that first minute and a half in order to solve this problem with the LCT.

Trefor Bazett - 3203 video solution

video by Trefor Bazett

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\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{1}{\sqrt{n^2+2}} } }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{1}{\sqrt{n^2+2}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

The Organic Chemistry Tutor - 3201 video solution

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n^2-4}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=3}^{\infty}{ \frac{1}{\sqrt{n^2-4}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Dr Chris Tisdell - 136 video solution

video by Dr Chris Tisdell

Final Answer

The series diverges by the Limit Comparison Test.

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2}{n^5+8} } }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^2}{n^5+8} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

The Organic Chemistry Tutor - 3200 video solution

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3n+1}{(2n-1)^4} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3n+1}{(2n-1)^4} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Dr Chris Tisdell - 138 video solution

video by Dr Chris Tisdell

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{6n-1}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{\sqrt{6n-1}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Dr Chris Tisdell - 139 video solution

video by Dr Chris Tisdell

Final Answer

The series diverges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{10n^2}{n^3-1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{10n^2}{n^3-1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Krista King Math - 140 video solution

video by Krista King Math

Final Answer

The series diverges by the Limit Comparison Test.

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n+5} } }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{3^n+5} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

The Organic Chemistry Tutor - 3202 video solution

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{2^n-1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Trefor Bazett - 3204 video solution

video by Trefor Bazett

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{2^n - 7} } }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{2^n - 7} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Ludus - 3215 video solution

video by Ludus

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{4^n-3} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{3}{4^n-3} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

MIP4U - 141 video solution

video by MIP4U

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{\sqrt{n^2-1}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{\sqrt{n^2-1}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

MIP4U - 142 video solution

video by MIP4U

Final Answer

The series diverges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n^2-1}{3n^5+2n+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Ludus - 3213 video solution

video by Ludus

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^4+2n^2-1}{6n^6+4n^4} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^4+2n^2-1}{6n^6+4n^4} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

MIP4U - 143 video solution

video by MIP4U

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{2/3}+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^{2/3}+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

Dr Phil Clark - 144 video solution

video by Dr Phil Clark

Final Answer

The series diverges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3-4} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2}{n^3-4} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 146 video solution

video by PatrickJMT

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{\sqrt{n}}{n-1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=2}^{\infty}{ \frac{\sqrt{n}}{n-1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 147 video solution

video by PatrickJMT

Final Answer

The series diverges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{2n^2+3} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{5}{2n^2+3} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 149 video solution

video by PatrickJMT

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3+1}{n^4-n+5} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3+1}{n^4-n+5} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 150 video solution

video by PatrickJMT

Final Answer

The series diverges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{n+1}{3n^3(4n-1)}}}\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{n+1}{3n^3(4n-1)}}}\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 152 video solution

video by PatrickJMT

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3-n+14}{n^5+n^3+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n^3-n+14}{n^5+n^3+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 154 video solution

video by PatrickJMT

Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+4} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the limit comparison test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1}{n^2+4} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

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 practice problem 2440 on the practice problem page.

Final Answer

The series converges by the limit comparison test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2-1}{n^3+4}}}\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the limit comparison test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2-1}{n^3+4}}}\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

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.

For more detail and other potential solutions, see the practice problems page.

Final Answer

The series diverges by the limit comparison test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt{n+1}}{2n^2+n+1} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

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Intermediate

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{n+1}{n\sqrt{n}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{1+3^n}{4+2^n} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{2\sqrt{n+3}}{n+5}}}\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{\sum_{n=1}^{\infty}{\frac{2\sqrt{n+3}}{n+5}}}\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 151 video solution

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Final Answer

The series diverges by the Limit Comparison Test.

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{9^n}{3+10^n} } }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{9^n}{3+10^n} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{e^{1/n}}{n} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4n^2}{\sqrt{9n^4+1}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Divergence Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{4n^2}{\sqrt{9n^4+1}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 153 video solution

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Final Answer

The series diverges by the Divergence Test.

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n^3+7}{n^4\sin^2(n)} } }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{2n^3+7}{n^4\sin^2(n)} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

He chose to compare this series with \(\sum{2/n}\). You could have used \(\sum{1/n}\) or any constant multiple of \(\sum{1/n}\). We probably would have chosen \(\sum{1/n}\).

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Advanced

\(\displaystyle{ \sum_{n=1}^{\infty}{(e^{1/n}-1)} }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{(e^{1/n}-1)} }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^{3/2}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series converges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\ln(n)}{n^{3/2}} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

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Final Answer

The series converges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt[3]{n}+\ln(n^4)}{n} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \frac{\sqrt[3]{n}+\ln(n^4)}{n} } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

PatrickJMT - 155 video solution

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Final Answer

The series diverges by the Limit Comparison Test.

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

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \sin(1/n) } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Final Answer

The series diverges by the Limit Comparison Test.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \sin(1/n) } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

This problem is solved by two different instructors.

blackpenredpen - 156 video solution

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PatrickJMT - 156 video solution

video by PatrickJMT

Final Answer

The series diverges by the Limit Comparison Test.

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\(\displaystyle{ \sum_{n=1}^{\infty}{ \sin \left[ \frac{1}{n^2} \right] } }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \sin \left[ \frac{1}{n^2} \right] } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Solution

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\(\displaystyle{ \sum_{n=1}^{\infty}{ ( \sqrt[n]{2} - 1 ) } }\)

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ ( \sqrt[n]{2} - 1 ) } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Hint

Choose \( \sum{1/n} \) as your test series.

Problem Statement

Determine the convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ ( \sqrt[n]{2} - 1 ) } }\). If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

Hint

Choose \( \sum{1/n} \) as your test series.

Solution

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Really UNDERSTAND Calculus

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Topics You Need To Understand For This Page

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).

Related Topics and Links

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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Limit Comparison Test

What The Limit Comparison Test Says

When To Use The Limit Comparison Test

How To Choose A Test Series

Limit Comparison Test Proof

Practice

Practice Search

Practice Instructions

Unless otherwise instructed, determine the convergence or divergence of the following series using the limit comparison test (LCT), if possible. [Of course, if the LCT is inconclusive, use another test to determine convergence or divergence.]

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