Comparison test

#### 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 comparison test for convergence

1. If the infinite series   $\sum_{n=1}^\infty b_n$   converges, and $0 \le a_n \le b_n$ for all $n \gt N,$ where N is some fixed number, then   $\sum_{n=1}^\infty a_n$   also converges.

2. If the infinite series   $\sum_{n=1}^\infty b_n$   diverges, and $0 \le b_n \le a_n$ for all $n \gt N,$ where N is some fixed number, then   $\sum_{n=1}^\infty a_n$   also diverges.

#### 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 bn
2. Each of the terms, bn, is larger than the corresponding term (an) 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.

The logic behind the second part of the comparison test goes like this:

1. We have a series we already know is divergent, with terms bn.
2. Each of the terms, an, in the series of interest is larger than the corresponding term (bn) of the known (divergent) series.
3. Then the series a1 + a2 + a3 + ... must diverge because it's larger, term-by-term, than a known divergent series. In other words, if the series with terms bn diverges, then the an series diverges faster.

We get the known divergent series used for comparison from among the many divergent series we've seen. For example, it might be a divergent geometric series (r < 1) or a divergent p-series (p < 1).

### Example 1

Does the series   $\sum_{n = 1}^\infty \,\frac{1}{n!}$   converge?

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

$$\sum_{n = 1}^\infty \,\frac{1}{n^2}$$

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

$$\sum_{n = 1}^\infty \,\frac{1}{n}$$

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 an = 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.

Now it's worth going back to the first green box above. The comparison test states that the terms of our test series (1/n!) must only be less, term-by-term, than a known convergent series for n > N, where N is some integer. In this example, N = 3. For all n > 3, 1/n! < 1/n2, so the series converges. The extra stuff at the "front end" of the series (n = 1, 2, 3) is just a constant added on to a convergent series.

### Example 2

Does the series   $\sum_{n = 1}^\infty \, \frac{\sqrt{n}}{n^2 + 5}$   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:

$$\sum_{n = 1}^\infty \, \frac{\sqrt{n}}{n^2} = \sum_{n = 1}^\infty \frac{n^{1/2}}{n^2} = \sum_{n = 1}^\infty \,\frac{1}{n^{3/2}}$$

Now the series with terms an = 1/n3/2 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:

$$\frac{\sqrt{n}}{n^2 + 5} \lt \frac{1}{n^{3/2}}$$

Cross multiplication gives:

$$n^{3/2} n^{1/2} \lt n^2 + 5$$

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

$$n^2 \lt n^2 + 5$$

which reduces to

$$0 \lt 5$$

Because 0 < 5 is always true, our series is, term-by-term, smaller than the convergent p-series with an = 1/n3/2, so it converges.

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

Below is a graphical representation of that comparison out to n = 30. While our the size of the terms of our function get closer to those of the comparison function, our algebra proves that they never cross.

### Example 3

Does the series   $\sum_{n = 1}^\infty \,\frac{1}{n^2} cos \left(\frac{1}{n} \right)$   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.

That means that this function is never greater than the convergent p-series with terms an = 1/n2, so the series converges by the comparison test.

Easy peasy, lemon-squeezy.

### Example 4

Does the series   $\sum_{n = 1}^\infty \,\frac{n + 1}{\sqrt{n}}$   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 n1/2, and the sum will grow infinitely large. But let's do it by comparison with the divergent p-series with terms an = 1/n1/2.

First let's break the sum into two parts.

$$\sum_{n = 1}^\infty \, \frac{n}{\sqrt{n}} + \frac{1}{\sqrt{n}} = \sum_{n = 1}^\infty \, n^{1/2} + \frac{1}{n^{1/2}}$$

Now the second is the divergent p-series with terms an = 1/n1/2 so the n1/2 term just adds an ever-increasing amount to an already divergent series, therefore the series diverges.

#### A couple of cautions

There are a couple of things you should be aware of when using the comparison test.

1. If the terms of a series, taken term-by-term, are less than the terms of a known divergent series, we cannot conclude that it converges. It may not.

2. If the terms of a series, taken term-by-term, are greater than the terms of a known convergent series, we cannot conclude that it diverges. It may still converge.

### Practice problems

Determine whether these series converge using the comparison test.

 1 $$\sum_{n = 2}^\infty \, \frac{n^3}{n^4 - 1}$$ 2 $$\sum_{n = 1}^\infty \, \frac{e^{1/n}}{n}$$ 3 $$\sum_{n = 0}^\infty \, \frac{n + 5^n}{n + 6^n}$$
 4 $$\sum_{n = 0}^\infty \, \frac{2 + sin(n)}{10^n}$$ 5 $$\sum_{n = 1}^\infty \, \frac{4^{n + 1}}{3^n - 2}$$ 6 $$\sum_{n = 1}^\infty \, \frac{n - 1}{n^2 \sqrt{n}}$$

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