Topics You Need To Understand For This Page
infinite limits 
infinite series basics 
pseries 
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). 
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Limit Comparison Test 
on this page:
► theorem ► what the test says ► when to use the test ► how to choose a test series

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

Quick Summary and Notes
used to prove convergence  yes 
used to prove divergence  yes 
can be inconclusive  yes (case 4) 
\( \sum{a_n} \) is the series we are trying to determine the convergence or divergence of; \( \sum{t_n} \) is the test series. 

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 pseries that you know either converges or diverges.
Idea 2: Choose a pseries 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 

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

Instructions   Unless otherwise instructed, determine the convergence or divergence of the following series using the limit comparison test, if possible.
Practice A01 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2+1}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2+1}}}\) diverges by the Limit Comparison Test 
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.
\(\displaystyle{ \lim_{n \to \infty}{ \left( \frac{n}{n^2+1} \div \frac{1}{n} \right) } = \lim_{n \to \infty}{ \left( \frac{n}{n^2+1} \cdot \frac{n}{1} \right) } = \lim_{n \to \infty}{\frac{n^2}{n^2+1}} = 1 }\)
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 pseries with \(p=1\), which means the original series also diverges by the Limit Comparison Test.
The Direct Comparison Test and Integral Test also work.
Practice A01 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n}{n^2+1}}}\) diverges by the Limit Comparison Test 
Practice A02 
\(\displaystyle{\sum_{n=0}^{\infty}{\frac{3}{n^2+1}}}\) 


\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{n^2+1} } }\) converges by the Limit Comparison Test 
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.
\(\displaystyle{ \lim_{n \to \infty}{ \left( \frac{3}{n^2+1} \div \frac{1}{n^2} \right) } =
\lim_{n \to \infty}{ \left( \frac{3}{n^2+1} \cdot \frac{n^2}{1} \right) } = \lim_{n \to \infty}{\frac{3n^2}{n^2+1}} = 3 }\)
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 pseries 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.
Practice A02 Final Answer 
\(\displaystyle{ \sum_{n=0}^{\infty}{ \frac{3}{n^2+1} } }\) converges by the Limit Comparison Test 
Practice A03 
\(\displaystyle{\sum_{n=0}^{\infty}{\frac{5n+2}{n^3+1}}}\) 


\(\displaystyle{ \sum_{n=0}^{\infty}{\frac{5n+2}{n^3+1}} }\) converges by the Limit Comparison Test 
Practice A03 Final Answer 
\(\displaystyle{ \sum_{n=0}^{\infty}{\frac{5n+2}{n^3+1}} }\) converges by the Limit Comparison Test 
Practice A04 
\(\displaystyle{\sum_{n=3}^{\infty}{\frac{1}{\sqrt{n^24}}}}\) 


\(\displaystyle{ \sum_{n=3}^{\infty}{\frac{1}{\sqrt{n^24}}} }\) diverges by the Limit Comparison Test 
Practice A04 Final Answer 
\(\displaystyle{ \sum_{n=3}^{\infty}{\frac{1}{\sqrt{n^24}}} }\) diverges by the Limit Comparison Test 
Practice A05 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{3n+1}{(2n1)^4}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{3n+1}{(2n1)^4}}}\) converges by the Limit Comparison Test 
Practice A05 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{3n+1}{(2n1)^4}}}\) converges by the Limit Comparison Test 
Practice A06 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{\sqrt{6n1}}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{\sqrt{6n1}}}}\) diverges by the Limit Comparison Test 
Practice A06 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{\sqrt{6n1}}}}\) diverges by the Limit Comparison Test 
Practice A07 
\(\displaystyle{\sum_{n=2}^{\infty}{\frac{10n^2}{n^31}}}\) 


\(\displaystyle{\sum_{n=2}^{\infty}{\frac{10n^2}{n^31}}}\) diverges by the Limit Comparison Test 
Practice A07 Final Answer 
\(\displaystyle{\sum_{n=2}^{\infty}{\frac{10n^2}{n^31}}}\) diverges by the Limit Comparison Test 
Practice A08 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{3}{4^n3}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{3}{4^n3}}}\) converges by the Limit Comparison Test 
Practice A08 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{3}{4^n3}}}\) converges by the Limit Comparison Test 
Practice A09 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{5}{\sqrt{n^21}}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{5}{\sqrt{n^21}}} }\) diverges by the Limit Comparison Test 
Practice A09 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{5}{\sqrt{n^21}}} }\) diverges by the Limit Comparison Test 
Practice A10 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^4+2n^21}{6n^6+4n^4}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^4+2n^21}{6n^6+4n^4}}}\) converges by the Limit Comparison Test 
Practice A10 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^4+2n^21}{6n^6+4n^4}}}\) converges by the Limit Comparison Test 
Practice A11 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^{2/3}+1}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^{2/3}+1}}}\) diverges by the Limit Comparison Test 
Practice A11 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^{2/3}+1}}}\) diverges by the Limit Comparison Test 
Practice A12 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{2}{n^34}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{2}{n^34}}}\) converges by the Limit Comparison Test 
Practice A12 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{2}{n^34}}}\) converges by the Limit Comparison Test 
Practice A13 
\(\displaystyle{\sum_{n=2}^{\infty}{\frac{\sqrt{n}}{n1}}}\) 


\(\displaystyle{\sum_{n=2}^{\infty}{\frac{\sqrt{n}}{n1}}}\) diverges by the Limit Comparison Test 
Practice A13 Final Answer 
\(\displaystyle{\sum_{n=2}^{\infty}{\frac{\sqrt{n}}{n1}}}\) diverges by the Limit Comparison Test 
Practice A14 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{5}{2n^2+3}}}\) 


The series converges by the Limit Comparison Test. 
Practice A14 Final Answer 
The series converges by the Limit Comparison Test. 
Practice A15 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^3+1}{n^4n+5}}}\) 


The series diverges by the Limit Comparison Test. 
Practice A15 Final Answer 
The series diverges by the Limit Comparison Test. 
Practice A16 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n+1}{3n^3(4n1)}}}\) 


The series converges by the Limit Comparison Test. 
Practice A16 Final Answer 
The series converges by the Limit Comparison Test. 
Practice A17 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{n^3n+14}{n^5+n^3+1}}}\) 


The series converges by the Limit Comparison Test. 
Practice A17 Final Answer 
The series converges by the Limit Comparison Test. 
Practice B01 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{2\sqrt{n+3}}{n+5}}}\) 


The series diverges by the Limit Comparison Test. 
Practice B01 Final Answer 
The series diverges by the Limit Comparison Test. 
Practice B02 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{4n^2}{\sqrt{9n^4+1}}}}\) 


The series diverges by the nthTerm Test. 
Practice B02 Final Answer 
The series diverges by the nthTerm Test. 
Practice C01 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\ln(n)}{n^{3/2}}}}\) 


\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\ln(n)}{n^{3/2}}}}\) converges by the Limit Comparison Test 
Practice C01 Final Answer 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\ln(n)}{n^{3/2}}}}\) converges by the Limit Comparison Test 
Practice C02 
\(\displaystyle{\sum_{n=1}^{\infty}{\frac{\sqrt[3]{n}+\ln(n^4)}{n}}}\) 


The series diverges by the Limit Comparison Test. 
Practice C02 Final Answer 
The series diverges by the Limit Comparison Test. 
Practice C03 
\(\displaystyle{\sum_{n=1}^{\infty}{\sin(1/n)}}\) 


The series diverges by the Limit Comparison Test. 
Practice C03 Final Answer 
The series diverges by the Limit Comparison Test. 