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:

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**).

**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/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 (

**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 (

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:

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.

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

Easy peasy, lemon-squeezy.

**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

First let's break the sum into two parts.

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.

Determine whether these series converg using the comparison test.

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