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:

- 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.
- 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.

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

- We have a series we already know is convergent, with terms
**b**_{n} **Each**of the terms,**b**, is larger than the corresponding term (_{n}**a**) of the series of interest._{n}- Then the series
**a**must converge because it's smaller, term-by-term, than a known convergent series._{1}+ a_{2}+ a_{3}+ ...

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:

- We have a series we already know is divergent, with terms
**b**._{n} **Each**of the terms,**a**, in the series of interest is larger than the corresponding term (_{n}**b**) of the known (divergent) series._{n}- Then the series
**a**must diverge because it's larger, term-by-term, than a known divergent series. In other words, if the series with terms_{1}+ a_{2}+ a_{3}+ ...**b**diverges, then the_{n}**a**series diverges_{n}*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**).

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

**Solution**

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

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/n ^{2}**, then the series converges by the comparison test.

We know that for **n ≥ 4**, **n! > n ^{2}**, so each term of the series with

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/n ^{2}**, so the series converges. The extra stuff at the "front end" of the series (

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

**Solution****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 **a _{n} = 1/n^{3/2}** is a convergent p-series (

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

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 **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:

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.

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

**Solution****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 **a _{n} = 1/n^{2}**, so the series converges by the comparison test.

__Easy peasy, lemon-squeezy.__

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

**Solution****n ^{1/2}**, and the sum will grow infinitely large. But let's do it by comparison with the divergent p-series with terms

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

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

- 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. - 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.

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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