(ST4) Limit Comparison Test#

In this lesson we are going to see how to:

  • use the Limit Comparison Test to show that a series is either convergent or divergent.

Review Video#

The Test#

Limit Comparison Test

Suppose \(\sum a_n\) and \(\sum b_n\) are series with positive terms where:

\[ \lim_{n\to\infty} \dfrac{a_n}{b_n}=c \]

If \(0<c<\infty\) then either both series converge or both series diverge.

  • \(\sum b_n\) is the series with known convergence / divergence

  • \(\sum a_n\) is the series we are testing.

Tip

As usual, we can relax the condition of the Limit Comparison Test so that the terms are eventually all positive.

Comparable Series?#

Similar to the original Comparison Test, we normally choose \(\sum b_n\) to be either a geometric series or \(p\)-series:

\(p\) Series
\[ \sum_{n=1}^{\infty} \dfrac{1}{n^p} \]

  • diverges when \(p\leq 1\)

  • converges when \(p > 1\)

Geometric Series
\[ \sum_{n=0}^{\infty} ar^n \]

  • converges when \(|r|<1\)

  • diverges when \(|r|\geq 1\)

Example 1#

Use the limit comparison test to determine if the series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{1}{3^n-1} \]

(Click to see the steps.)

We first need to decide which series we are going to compare this to. We notice that it looks like the geometric series:

\[ \sum_{n=1}^{\infty} \dfrac{1}{3^n} \]

We know this is convergent, since this is a geometric series with \(r= \tfrac{1}{3}\) and \(|r| < 1\).

So this is going to be our choice for \(\sum b_n\).

We are comparing the two series \(\sum a_n\) and \(\sum b_n\) where:

\[ a_n = \dfrac{1}{3^n-1}\qquad \text{and}\qquad b_n = \dfrac{1}{3^n} \]

  1. Positive Terms? Yes, every term in \(a_n\) and \(b_n\) is positive for \(n\geq 1\). Note that \(3^n-1 > 0\) for \(n\geq 1\).

  2. Limit: We next need to compute the limit:

\[\begin{split} \lim_{n\to \infty} \dfrac{a_n}{b_n} & = \lim_{n\to \infty} a_n \cdot \dfrac{1}{b_n} \\[10pt] & = \lim_{n\to \infty} \dfrac{1}{3^n-1} \cdot \dfrac{3^n}{1} \\[10pt] & = \lim_{n\to \infty} \dfrac{3^n}{3^n-1} \\[10pt] & = \lim_{n\to \infty} \dfrac{\tfrac{3^n}{3^n}}{\tfrac{3^n}{3^n}-\tfrac{1}{3^n}} \\[10pt] & = \lim_{n\to \infty} \dfrac{1}{1-\tfrac{1}{3^n}} \\[10pt] & = \dfrac{1}{1-0} \\[10pt] & = 1 \\[10pt] \end{split}\]

Since this limit exists and is greater than \(0\), we are able to use the Limit Comparison Test.

The two series we compared:

\[ \sum a_n = \sum_{n=1}^{\infty} \dfrac{1}{3^n-1}\qquad \text{and}\qquad \sum b_n = \sum_{n=1}^{\infty} \dfrac{1}{3^n} \]

satisfy the necessary conditions for \(n\geq 1\) (including the limit existing and being greater than 0). Since \(\sum b_n\) converges, we conclude that our original series \(\sum a_n\) converges as well by the Limit Comparison Test.

Example 2#

Determine if the following series is convergent or divergent.

\[ \sum_{n=1}^{\infty} \dfrac{\sqrt{2n^3+1}}{5n^3-2n} \]

(Click to see the steps.)

We first need to decide which series we are going to compare this to. We do this, by only using the dominant terms in the numerator and denominator:

\[ \sum_{n=1}^{\infty} \dfrac{\sqrt{2n^3}}{5n^3} = \sum_{n=1}^{\infty} \dfrac{\sqrt{2}n^{3/2}}{5n^3} = \sum_{n=1}^{\infty} \dfrac{\sqrt{2}}{5}\cdot \dfrac{1}{n^{3/2}} = \dfrac{\sqrt{2}}{5}\cdot\sum_{n=1}^{\infty} \dfrac{1}{n^{3/2}} \]

We know this is convergent, since this is a constant multiple of a \(p\)-series with \(p=3/2>1\).

So this is going to be our choice for \(\sum b_n\).

We are comparing the two series \(\sum a_n\) and \(\sum b_n\) where:

\[ a_n = \dfrac{\sqrt{2n^3+1}}{5n^3-2n} \qquad \text{and}\qquad b_n = \dfrac{\sqrt{2n^3}}{5n^3} \]

  1. Positive Terms? Yes, every term in \(a_n\) and \(b_n\) is positive for \(n\geq 1\). Note that \( 5n^3-2n > 0\) for \(n\geq 1\).

  2. Limit: We next need to compute the limit:

\[\begin{split} \lim_{n\to \infty} \dfrac{a_n}{b_n} & = \lim_{n\to \infty} a_n \cdot \dfrac{1}{b_n} \\[10pt] & = \lim_{n\to \infty} \dfrac{\sqrt{2n^3+1}}{5n^3-2n} \cdot \dfrac{5n^3}{\sqrt{2n^3}} \\[10pt] & = \lim_{n\to \infty} \dfrac{\sqrt{2n^3+1}}{\sqrt{2n^3}}\cdot \dfrac{5n^3}{5n^3-2n} \\[10pt] & = \lim_{n\to \infty} \dfrac{\sqrt{2n^3+1}}{\sqrt{2n^3}}\cdot \lim_{n\to \infty} \dfrac{5n^3}{5n^3-2n} \\[10pt] & = \lim_{n\to \infty} \sqrt{\dfrac{2n^3+1}{2n^3}} \cdot \lim_{n\to \infty} \dfrac{5n^3}{5n^3-2n} \\[10pt] & = \sqrt{\lim_{n\to \infty} \dfrac{2n^3+1}{2n^3}}\cdot \lim_{n\to \infty} \dfrac{5n^3}{5n^3-2n} \\[10pt] & = \sqrt{\lim_{n\to \infty} \dfrac{\tfrac{2n^3}{2n^3}+\tfrac{1}{2n^3}}{\tfrac{2n^3}{2n^3}}}\cdot \lim_{n\to \infty} \dfrac{\tfrac{5n^3}{5n^3}}{\tfrac{5n^3}{5n^3}-\tfrac{2n}{5n^3}} \\[10pt] & = \sqrt{\lim_{n\to \infty} \dfrac{1+\tfrac{1}{2n^3}}{1 }}\cdot \lim_{n\to \infty} \dfrac{1}{1-\tfrac{2}{5n^2}} \\[10pt] & = \sqrt{ \dfrac{1+0}{1 }}\cdot \dfrac{1}{1-0} \\[10pt] & = 1 \\[10pt] \end{split}\]

Since this limit exists and is greater than \(0\), we are able to use the Limit Comparison Test.

The two series we compared:

\[ \sum a_n = \sum_{n=1}^{\infty} \dfrac{\sqrt{2n^3+1}}{5n^3-2n}\qquad \text{and}\qquad \sum b_n = \sum_{n=1}^{\infty} \dfrac{\sqrt{2n^3}}{5n^3} \]

satisfy the necessary conditions for \(n\geq 1\) (including the limit existing and being greater than 0). Since \(\sum b_n\) converges, we conclude that our original series \(\sum a_n\) converges as well by the Limit Comparison Test.