(ST3) Direct Comparison Test#

In this lesson we are going to see how to:

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

Review Video#

Motivation#

The idea behind the comparison tests is to:

  • compare a series with unknown convergence

  • to a series with known convergence.

Unknown Convergence
\[ \sum_{n=1}^{\infty}\dfrac{1}{3^n+1} \]
Known Convergence
\[ \sum_{n=1}^{\infty}\dfrac{1}{3^n} \]

The Test#

Comparison Test

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

  • If \(\sum b_n\) is convergent and \(a_n\leq b_n\) for all \(n\), then \(\sum a_n \) is convergent.

  • If \(\sum b_n\) is divegent and \(b_n\leq a_n\) for all \(n\), then \(\sum a_n \) is divergent.

In each of these:

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

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

What to Compare?#

When we pick \(\sum b_n\), its convergence or divergence must be known! So we usually try to use either a geometric series or \(p\)-series:

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

  • converges when \(|r|<1\)

  • diverges when \(|r|\geq 1\)

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

  • diverges when \(p\leq 1\)

  • converges when \(p > 1\)

Starting Index Value#

Tip

As usual, we can relax the conditions with our starting index value to include other values besides \(n=1\), and the Comparison Test still applies.

This means we can relax the conditions so that the two series we’re comparing only need to eventually:

  • have all positive terms and

  • satisfy the necessary inequality.

Eventually - we can find some integer \(N\) where the terms in the series satisfy the condition for all \(n\geq N\).

Example 1#

Use the 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 and there is no subtraction.

  2. Inequality: Since \(\sum b_n\) is convergent, we suspect \(\sum a_n\) is convergent. This means we need to show the inequality \(a_n\leq b_n\).

In this example, we start with an inequality that is clearly true, and then produce the desired result:

\[\begin{split} & 3^n &< 3^n +1 \\[10pt] \iff & \dfrac{1}{3^n} &> \dfrac{1}{3^n+1} \\[10pt] \iff & b_n &> a_n \\[10pt] \end{split}\]

The two series we compared:

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

both satisfy the necessary conditions for \(n\geq 1\). And since \(\sum b_n\) converges, we conclude that our original series \(\sum a_n\) converges as well by the Comparison Test.

Example 2#

Determine if the following series is convergent or divergent.

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

(Click to see the steps.)

We first need to find a comparable series.

By looking at the dominant term in the denominator (and numerator if necessary), we notice that this series looks like the \(p\)-series:

\[ \sum_{n=1}^{\infty} \dfrac{3}{2n^2} \]

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

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:

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

  1. Positive Terms? Yes, every term in \(a_n\) and \(b_n\) is positive and there is no subtraction.

  2. Inequality: Since \(\sum b_n\) is convergent, we suspect \(\sum a_n\) is convergent. This means we need to show the inequality \(a_n\leq b_n\).

In this example, we will start with our desired inequality and simplify the result until we get a statement that is clearly true. For \(n\geq 1\),

\[\begin{split} &\dfrac{3}{2n^2+5n+1} &< \dfrac{3}{2n^2} \\[10pt] \iff & \qquad \; 3(2n^2)&< 3(2n^2+5n+1) \\[10pt] \iff & \qquad \quad 6n^2 &< 6n^2+15n+3 \\[10pt] \iff & \qquad \quad \; 0 &< 15n+3 \\[10pt] \end{split}\]

This last inequality is clearly true for \(n\geq 1\), which means the original inequality must also be true.

The two series we compared:

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

both satisfy the necessary conditions for \(n\geq 1\). And since \(\sum b_n\) converges, we conclude that our original series \(\sum a_n\) converges as well by the Comparison Test.

Example 3#

Determine if the following series is convergent or divergent.

\[ \sum_{n=0}^{\infty} \dfrac{3^{n+1}}{2^n-5} \]

(Click to see the steps.)

We first need to find a comparable series.

By looking at the dominant term in the denominator (and numerator if necessary), we notice that this series looks like the geometric series:

\[ \sum_{n=0}^{\infty} \dfrac{3^{n+1}}{2^n} = \sum_{n=0}^{\infty} \dfrac{3\cdot 3^{n}}{2^n} = \sum_{n=0}^{\infty} 3\cdot \left(\dfrac{3}{2}\right)^n \]

We know this is divergent, since this is a geometric series with \(r= \tfrac{3}{2}\) and \(|r|\geq 1\).

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:

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

  1. Positive Terms? Almost, once \(n\geq 3\) then every \(a_n\) and \(b_n\) term is positive.

  2. Inequality: Since \(\sum b_n\) is divergent, we suspect \(\sum a_n\) is divergent. This means we need to show the inequality \(b_n\leq a_n\).

In this example, we will start with our desired inequality and simplify the result until we get a statement that is clearly true. For \(n\geq 3\),

\[\begin{split} & \quad \dfrac{3^{n+1}}{2^n} &< \dfrac{3^{n+1}}{2^n-5} \\[10pt] \iff & \quad \dfrac{1}{2^n} &< \dfrac{1}{2^n-5} \\[10pt] \iff & 2^n-5 &< 2^n \\[10pt] \end{split}\]

This last inequality is clearly true, which means the original inequality must also be true.

The two series we compared:

\[ \sum a_n = \sum_{n=0}^{\infty} \dfrac{3^{n+1}}{2^n-5} \qquad \text{and}\qquad \sum b_n = \sum_{n=1}^{\infty} \dfrac{3^{n+1}}{2^n} \]

both satisfy the necessary conditions for \(n\geq 3\). And since \(\sum b_n\) diverges, we conclude that our original series \(\sum a_n\) diverges as well by the Comparison Test.