Calculus/Limit Comparison Test

Limit Comparison Test

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.

Limit Comparison Test Definition

For the two series $\textstyle {\sum {a_{n}}}$ and $\textstyle {\sum {t_{n}}}$ where $a_{n}>0$ and $t_{n}>0$ , we calculate $\displaystyle {\lim _{n\to \infty }{\left[{\frac {a_{n}}{t_{n}}}\right]}=L}$ .

When $L$ is finite and positive, the two series either both converge or both diverge.
When $L=0$ and $\textstyle {\sum {t_{n}}}$ converges, then $\textstyle {\sum {a_{n}}}$ also converges.
When $L=\infty$ and $\textstyle {\sum {t_{n}}}$ diverges, then $\textstyle {\sum {a_{n}}}$ also diverges.
If the limit does not exist then the test is inconclusive.

Limit Comparison Test Quick Notes

used to prove convergence	yes
used to prove divergence	yes
can be inconclusive	yes

Notice that we do not specify the $n$ -values on the sum. This is common in calculus and it just means that, for this test, it doesn't matter where the series starts (but it always 'ends' at infinity, since this is an infinite series).
$\textstyle {\sum {a_{n}}}$ is the series of which we are trying to determine convergence or divergence and it is given in the problem statement.
$\textstyle {\sum {t_{n}}}$ is the test series that you choose.

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 $\textstyle {\sum {a_{n}}}$ and the test series that you know whether it converges or diverges is $\textstyle {\sum {t_{n}}}$ . The limit $L$ has to be calculated 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 $n\geq 1$ to replace $\ln(n)$ with $n$ or use $\ln(n)\geq 1$ for $n\geq 3$ .

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

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 who are interested.

Proof of the Limit Comparison Test

Video Recommendations

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.

Watching these two video clips will give you a better feel for the limit comparison test. Both clips are short and to the point.

Series, comparison + ratio tests

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

Limit Comparison Test Practice Problems

Practice Problems with Written Solutions

Determine the convergence or divergence of the following series. If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

1. $\sum _{n=1}^{\infty }{\frac {n^{2}-1}{n^{3}+4}}$

Hint

Choose

t_{n}=1/n

Choose

t_{n}=1/n

Answer

The series diverges by the limit comparison test.

Solution

Use the test series

\textstyle {\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.

Use the test series

\textstyle {\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.

2. $\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n^{2}+4}}}$

Hint

Choose

t_{n}=1/n^{2}

Choose

t_{n}=1/n^{2}

Answer

The series converges by the limit comparison test.

Solution

Compare to

\textstyle {\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 $0\leq \lim _{n\to \infty }{\frac {a_{n}}{t_{n}}}=1<\infty$ and $\textstyle {\sum {t_{n}}}$ converges, the series also converges.

The limit comparison test is one of the best tests for this series.

Compare to

\textstyle {\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 $0\leq \lim _{n\to \infty }{\frac {a_{n}}{t_{n}}}=1<\infty$ and $\textstyle {\sum {t_{n}}}$ converges, the series also converges.

The limit comparison test is one of the best tests for this series.

3. $\displaystyle {\sum _{n=0}^{\infty }{\frac {3}{n^{2}+1}}}$

Hint

Choose

t_{n}=1/n^{2}

Choose

t_{n}=1/n^{2}

Answer

The series converges by the Limit Comparison Test.

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 $\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$ .

${\begin{array}{rcl}\displaystyle {\lim _{n\to \infty }{\frac {a_{n}}{t_{n}}}}&=&\displaystyle {\lim _{n\to \infty }{\left({\frac {3}{n^{2}+1}}\div {\frac {1}{n^{2}}}\right)}}\\&=&\displaystyle {\lim _{n\to \infty }{\left({\frac {3}{n^{2}+1}}\cdot {\frac {n^{2}}{1}}\right)}}\\&=&\displaystyle {\lim _{n\to \infty }{\frac {3n^{2}}{n^{2}+1}}}\\&=&\displaystyle {\lim _{n\to \infty }{\frac {3}{1+1/n^{2}}}=3}\\\end{array}}$

Since the limit is positive and finite, the two series either both converge or both diverge. The series $\textstyle {\sum {1/n^{2}}}$ 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.

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 $\displaystyle {\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$ .

${\begin{array}{rcl}\displaystyle {\lim _{n\to \infty }{\frac {a_{n}}{t_{n}}}}&=&\displaystyle {\lim _{n\to \infty }{\left({\frac {3}{n^{2}+1}}\div {\frac {1}{n^{2}}}\right)}}\\&=&\displaystyle {\lim _{n\to \infty }{\left({\frac {3}{n^{2}+1}}\cdot {\frac {n^{2}}{1}}\right)}}\\&=&\displaystyle {\lim _{n\to \infty }{\frac {3n^{2}}{n^{2}+1}}}\\&=&\displaystyle {\lim _{n\to \infty }{\frac {3}{1+1/n^{2}}}=3}\\\end{array}}$

Since the limit is positive and finite, the two series either both converge or both diverge. The series $\textstyle {\sum {1/n^{2}}}$ 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.

4. $\displaystyle {\sum _{n=1}^{\infty }{\frac {n}{n^{2}+1}}}$

Hint

Choose

t_{n}=1/n

Choose

t_{n}=1/n

Answer

The series diverges by the Limit Comparison Test.

Video Solution

link to solution video

Written 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}}}

.

${\begin{array}{rcl}\displaystyle {\lim _{n\to \infty }{\frac {a_{n}}{t_{n}}}}&=&\displaystyle {\lim _{n\to \infty }{\left({\frac {n}{n^{2}+1}}\div {\frac {1}{n}}\right)}}\\&=&\displaystyle {\lim _{n\to \infty }{\left({\frac {n}{n^{2}+1}}\cdot {\frac {n}{1}}\right)}}\\&=&\displaystyle {\lim _{n\to \infty }{\frac {n^{2}}{n^{2}+1}}=1}\end{array}}$

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

The Direct Comparison Test and Integral Test also work.

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}}}

.

${\begin{array}{rcl}\displaystyle {\lim _{n\to \infty }{\frac {a_{n}}{t_{n}}}}&=&\displaystyle {\lim _{n\to \infty }{\left({\frac {n}{n^{2}+1}}\div {\frac {1}{n}}\right)}}\\&=&\displaystyle {\lim _{n\to \infty }{\left({\frac {n}{n^{2}+1}}\cdot {\frac {n}{1}}\right)}}\\&=&\displaystyle {\lim _{n\to \infty }{\frac {n^{2}}{n^{2}+1}}=1}\end{array}}$

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

The Direct Comparison Test and Integral Test also work.

Practice Problems with Video Solutions

Determine the convergence or divergence of the following series. If possible, use the Limit Comparison Test. If the LCT is inconclusive, use another test to determine convergence or divergence.

1	$\sum _{n=0}^{\infty }{\frac {5n+2}{n^{3}+1}}$	answer converges converges	solution	2	$\sum _{n=3}^{\infty }{\frac {1}{\sqrt {n^{2}-4}}}$	answer diverges diverges	solution
3	$\sum _{n=1}^{\infty }{\frac {\ln(n)}{n^{3/2}}}$	answer converges converges	solution	4	$\sum _{n=1}^{\infty }{\frac {3n+1}{(2n-1)^{4}}}$	answer converges converges	solution
5	$\sum _{n=1}^{\infty }{\frac {1}{\sqrt {6n-1}}}$	answer diverges diverges	solution	6	$\sum _{n=2}^{\infty }{\frac {10n^{2}}{n^{3}-1}}$	answer diverges diverges	solution
7	$\sum _{n=1}^{\infty }{\frac {5}{\sqrt {n^{2}-1}}}$	answer diverges diverges	solution	8	$\sum _{n=1}^{\infty }{\frac {n^{4}+2n^{2}-1}{6n^{6}+4n^{4}}}$	answer converges converges	solution
9	$\sum _{n=1}^{\infty }{\frac {1}{n^{2/3}+1}}$	answer diverges diverges	solution	10	$\sum _{n=1}^{\infty }{\frac {2}{n^{3}-4}}$	answer converges converges	solution
11	$\sum _{n=2}^{\infty }{\frac {\sqrt {n}}{n-1}}$	answer diverges diverges	solution	12	$\sum _{n=1}^{\infty }{\frac {5}{2n^{2}+3}}$	answer converges converges	solution
13	$\sum _{n=1}^{\infty }{\frac {n^{3}+1}{n^{4}-n+5}}$	answer diverges diverges	solution	14	$\sum _{n=1}^{\infty }{\frac {2{\sqrt {n+3}}}{n+5}}$	answer diverges diverges	solution