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.