Deciding convergence by comparison to a series already known to converge or diverge

The comparison test is an easy test for convergence or divergence when we know that, term by term, the terms of a test series are either greater than or less than those of our series of interest.

Here is a statement of the test in two parts, then we'll do some examples to illustrate it:

The logic of the test

Here is the logic behind the first statement of the comparison test:

1. We have a series we already know is convergent, with terms b_n .
2. Each of the terms, b_n, is larger than the corresponding term (a_n) of the series of interest.
3. Then the series a₁ + a₂ + a₃ + ... must converge because it's smaller, term-by-term, than a known convergent series.

We get the convergent series used for comparison from our growing set of convergent series. For example, it might be a convergent geometric series or a convergent p-series.

Example 1:   Does the series     converge?

Solution: When we look at this series, it's how rapidly the denominator grows that's important. We know that the series

,

the harmonic series, diverges, but that the p-series

converges. Now if each term of the series of interest is greater than 1/n for n > 1 then the series diverges, and if each term is less than 1/n², then the series converges by the comparison test.

We know that for n ≥ 4, n! > n², so each term of the series with a_n = 1/n! is less than the corresponding term of our convergent p-series, therefore the series converges. Here's a graphical look at these three series, 1/n in green, 1/n² in magenta and 1/n! in blue. Notice that for n ≥ 4, 1/n! is smaller, term-by-term, than 1/n².

Example 2:   Does the series     converge?

Solution: For a series like this, it's helpful to reduce it by approximation. In the first approximation, the 5 in the denominator doesn't really matter that much for large n, so let's drop it to make an approximation of the series:

Now the series with terms a_n = 1/n^3/2 is a convergent p-series (p = 3/2 > 1). That suggests that we can show that each of the terms of our series is smaller than the corresponding term of the p-series with p = 3/2. First we set up the trial inequality:

Cross multiplication gives:

Using the laws of exponents on the left, we get:

which reduces to

Because 0 < 5 is always true, our series is, term-by-term, smaller than the convergent p-series with a_n = 1/n^3/2, so it converges.

Here (right column, top) is a table of the first ten terms of our series and the comparison series:

Example 3:   Does the series     converge?

Solution: This series contains an oscillating factor, cos(1/n). Sometimes that factor is positive and sometimes it's negative. Now 1/n approaches zero as n gets large, so the cosine part approaches 1. The important thing is that it won't ever exceed 1.

Example 4:   Does the series     converge?

Solution: It's not too difficult to see that this series has to diverge. If we drop the 1 in the numerator, each term is just n^1/2, and the sum will grow infinitely large. But let's do it by comparison with the divergent p-series with terms a_n = 1/n^1/2.

First let's break the sum into two parts.

Practice problems

Determine whether these series converg using the comparison test.

You can help

This site is a one-person operation and a labor of love. If you can manage it, I'd appreciate anything you could give to help defray the cost of keeping up the domain name, server, search service and my time.