Direct Comparison Test

The direct comparison test compares a series with positive terms to a series you already know, term by term. If 0 \le a_n \le b_n and \sum b_n converges, then \sum a_n converges. If a_n \ge b_n \ge 0 and \sum b_n diverges, then \sum a_n diverges.

Before this lesson: p-Series, Infinite Geometric Series, Integral Test and Comparison Test for Improper Integrals

Why comparing works

When every term is positive, the partial sums only go up. A series like that can do only two things: level off at a finite number, or grow without bound.

Now suppose each term of your series is smaller than the matching term of a series whose total is finite. Your running total stays below that finite total, so it cannot grow without bound. It must level off, so it converges.

It works the other way too. If each term is bigger than the matching term of a series that grows without bound, your running total is pushed up without bound as well.

The direct comparison test

Suppose a_n and b_n are both nonnegative for all n (or from some point on).

\text{If } a_n \le b_n \text{ and } \sum b_n \text{ converges, then } \sum a_n \text{ converges.}

\text{If } a_n \ge b_n \text{ and } \sum b_n \text{ diverges, then } \sum a_n \text{ diverges.}

Two cases give no answer. Being smaller than a divergent series tells you nothing. Being bigger than a convergent series tells you nothing. The inequality has to point the right way.

Good series to compare with. Use a p-series \sum 1/n^p (converges when p > 1) or a geometric series \sum r^n (converges when |r| < 1).

How to use the direct comparison test

1. Check that the terms are nonnegative.
2. Find the part of the term that matters most for large n. Drop the smaller pieces to guess a p-series or geometric series that behaves the same way.
3. Decide whether you expect convergence or divergence, then prove the inequality in the matching direction.
4. State the conclusion. If the inequality goes the wrong way, try the limit comparison test instead.

Direct comparison test examples

Example 1: Adding to the denominator

Does \sum_{n=1}^{\infty} \dfrac{1}{n^2 + 5} converge?

\dfrac{1}{n^2 + 5} \gt 0

n^2 + 5 \gt n^2

\dfrac{1}{n^2 + 5} \lt \dfrac{1}{n^2}

\sum \dfrac{1}{n^2} \text{ converges}

Answer: the series converges by the direct comparison test, because its terms are smaller than those of a convergent series.

Example 2: Bigger than the harmonic series

Does \sum_{n=3}^{\infty} \dfrac{\ln n}{n} converge?

\ln n \ge \ln 3 \gt 1 \text{ for } n \ge 3

\dfrac{\ln n}{n} \gt \dfrac{1}{n}

\sum_{n=3}^{\infty} \dfrac{1}{n} \text{ diverges}

Answer: the series diverges by the direct comparison test, because its terms are bigger than those of a divergent series.

Example 3: Comparing with a geometric series

Does \sum_{n=1}^{\infty} \dfrac{3^n}{4^n + 2} converge?

4^n + 2 \gt 4^n

\dfrac{3^n}{4^n + 2} \lt \dfrac{3^n}{4^n}

\dfrac{3^n}{4^n} = \left(\dfrac{3}{4}\right)^n

\sum \left(\dfrac{3}{4}\right)^n \text{ converges}

Answer: the series converges by the direct comparison test.

Example 4: A bounded trig factor

Does \sum_{n=1}^{\infty} \dfrac{2 + \sin n}{n^2} converge?

-1 \le \sin n \le 1

1 \le 2 + \sin n \le 3

0 \lt \dfrac{2 + \sin n}{n^2} \le \dfrac{3}{n^2}

\sum \dfrac{3}{n^2} = 3\sum \dfrac{1}{n^2} \text{ converges}

Answer: the series converges. The integral test could not handle this series, because its terms go up and down, but a comparison works easily.

Example 5: Coins in a game (word problem)

A phone game gives a daily bonus of \dfrac{100}{n^2 + n + 1} coins on day n. Show that the total bonus stays below a fixed amount forever, and find such an amount.

\dfrac{100}{n^2 + n + 1} \lt \dfrac{100}{n^2}

\sum_{n=1}^{\infty} \dfrac{100}{n^2} = 100 \cdot \dfrac{\pi^2}{6}

100 \cdot \dfrac{\pi^2}{6} \approx 164.5

\sum_{n=1}^{\infty} \dfrac{100}{n^2 + n + 1} \lt 164.5

Answer: the total bonus converges and is always less than about 164.5 coins, no matter how long you play. (Its actual value is about 79.8 coins.)

Common mistakes

Related methods

When the inequality goes the wrong way

For \sum \frac{1}{n^2 - 1} (starting at n = 2), the terms are a little bigger than 1/n², so a direct comparison with 1/n² fails. The limit comparison test fixes this: it only asks whether the ratio of the terms approaches a positive number.

The same idea for integrals

The comparison test for improper integrals works the same way, with integrals in place of sums.

Practice problems

1. Does \sum_{n=1}^{\infty} \dfrac{1}{n^3 + 2} converge?
2. Does \sum_{n=1}^{\infty} \dfrac{1}{2^n + 7} converge?
3. Does \sum_{n=1}^{\infty} \dfrac{n + 1}{n^2} converge?
4. Does \sum_{n=1}^{\infty} \dfrac{\sin^2 n}{n^2} converge?
5. Does \sum_{n=1}^{\infty} \dfrac{1}{n + 1} converge?
6. Does \sum_{n=1}^{\infty} \dfrac{\ln n}{n^3} converge?
7. Does \sum_{n=1}^{\infty} \dfrac{1}{n!} converge?
8. Does \sum_{n=2}^{\infty} \dfrac{1}{\ln n} converge?
9. Does \sum_{n=1}^{\infty} \dfrac{\sqrt{n}}{n^2 + 1} converge?
10. A fundraiser collects \dfrac{50}{n \cdot 2^n} thousand dollars on day n. Show that the total is less than 50 thousand dollars.

\dfrac{1}{n^3 + 2} \lt \dfrac{1}{n^3}

\sum \dfrac{1}{n^3} \text{ converges}

\dfrac{1}{2^n + 7} \lt \dfrac{1}{2^n} = \left(\tfrac{1}{2}\right)^n

\sum \left(\tfrac{1}{2}\right)^n \text{ converges}

\dfrac{n + 1}{n^2} = \dfrac{1}{n} + \dfrac{1}{n^2}

\dfrac{1}{n} + \dfrac{1}{n^2} \gt \dfrac{1}{n}

\sum \dfrac{1}{n} \text{ diverges}

0 \le \sin^2 n \le 1

0 \le \dfrac{\sin^2 n}{n^2} \le \dfrac{1}{n^2}

\sum \dfrac{1}{n^2} \text{ converges}

n + 1 \le 2n \quad \text{for } n \ge 1

\dfrac{1}{n + 1} \ge \dfrac{1}{2n}

\sum \dfrac{1}{2n} = \dfrac{1}{2}\sum \dfrac{1}{n} \text{ diverges}

\ln n \lt n \quad \text{for } n \ge 1

0 \le \dfrac{\ln n}{n^3} \lt \dfrac{n}{n^3} = \dfrac{1}{n^2}

\sum \dfrac{1}{n^2} \text{ converges}

n! = 1 \cdot 2 \cdot 3 \cdots n \ge 1 \cdot 2 \cdot 2 \cdots 2 = 2^{n-1}

\dfrac{1}{n!} \le \dfrac{1}{2^{n-1}} = \left(\tfrac{1}{2}\right)^{n-1}

\sum \left(\tfrac{1}{2}\right)^{n-1} = 2

\ln n \lt n \quad \text{for } n \ge 2

\dfrac{1}{\ln n} \gt \dfrac{1}{n}

\sum_{n=2}^{\infty} \dfrac{1}{n} \text{ diverges}

n^2 + 1 \gt n^2

\dfrac{\sqrt{n}}{n^2 + 1} \lt \dfrac{\sqrt{n}}{n^2}

\dfrac{\sqrt{n}}{n^2} = \dfrac{n^{1/2}}{n^2} = \dfrac{1}{n^{3/2}}

\sum \dfrac{1}{n^{3/2}} \text{ converges}

\dfrac{50}{n \cdot 2^n} \le \dfrac{50}{2^n}

\sum_{n=1}^{\infty} \dfrac{50}{2^n} = \dfrac{25}{1 - \frac{1}{2}} = 50

\dfrac{50}{1 \cdot 2^1} = \dfrac{50}{2^1},\ \dfrac{50}{2 \cdot 2^2} \lt \dfrac{50}{2^2}

\sum_{n=1}^{\infty} \dfrac{50}{n \cdot 2^n} \lt 50

Direct comparison test FAQ

What is the direct comparison test?

It compares a series with nonnegative terms to a known series, term by term. Smaller than a convergent series means convergent; bigger than a divergent series means divergent.

What series should I compare with?

Use p-series like \sum 1/n^2 or \sum 1/n, and geometric series like \sum (1/2)^n. To choose, keep only the fastest-growing parts of the top and bottom of the term.

What if the inequality goes the wrong way?

Then the direct comparison test gives no answer. Try the limit comparison test, which compares the terms with a limit instead of an inequality.

Does the direct comparison test work for negative terms?

No. The terms must be nonnegative, at least from some point on. For series with mixed signs, look at absolute convergence or the alternating series test.

Summary

• For nonnegative terms: if a_n \le b_n and \sum b_n converges, then \sum a_n converges.
• If a_n \ge b_n and \sum b_n diverges, then \sum a_n diverges.
• Smaller than divergent, or bigger than convergent, tells you nothing.
• Compare with p-series and geometric series, chosen from the fastest-growing parts of the term.
• A comparison gives a bound on the sum, not its exact value.

Previous lesson: p-Series

Next lesson: Limit Comparison Test

Back to the Calculus 2 Index