# The Comparison and Limit Comparison Tests

You can often tell that a series converges or diverges by comparing it to a known series. I'll look first at situations where you can establish an inequality between the terms of two series.

Let , , be series with positive terms.

1. If for all k and converges, then converges.

2. If for all k and diverges, then diverges.

It's easy to see that these tests make sense. In the first case, the partial sums of are bounded above by . The partial sums increase, so they must converge.

In the second case, the partial sums are always bigger than the partial sums, but the partial sums go to . Hence, the partial sums go to as well.

Example. Determine whether converges or diverges.

The series has positive terms. In fact, I could use the Integral Test, but who would want to integrate ?

Instead, note that when k is large, the term should dominate. How does compare to ? Well, if you make the bottom smaller, the fraction gets bigger:

Now is a p-series with , so it converges. Hence, converges by comparison.

Example. Determine whether converges or diverges.

The series has positive terms. Since ,

The second inequality comes from the fact that making the bottom smaller makes the fraction bigger.

Now is a p-series with , so it converges. Hence, converges by comparison.

Example. Determine whether converges or diverges.

The series has positive terms. Since ,

converges, because it's a p-series with . Therefore, converges by direct comparison.

Example. Determine whether converges or diverges.

If you make the top smaller, the fraction gets smaller:

Notice how I avoided changing to k; I changed it to something which cancelled the radical on the bottom.

Now diverges --- it's harmonic! So diverges, by comparison.

Example. Comparison won't work if the inequalities go the wrong way. For example, consider . I'd like to compare this to , but if I make the bottom bigger (by adding 2), the fraction gets smaller:

It's true that is a convergent p-series ( ), but it's smaller than the given series. I can't draw a conclusion this way.

Nevertheless, is "close to" the given series in some sense. Limit Comparison will make precise the idea that one series is "close to" another, without having to worry about inequalities.

Let be a positive term series.

Let be a positive term series whose behavior is known. (You usually choose this series because it seems to be "close to" the given series.)

Look at the limiting ratio

1. If the limit is a finite positive number, then the two series behave in the same way. That is, if converges, then converges; if diverges, then diverges.

2. If the limit is 0 and converges, then converges.

3. If the limit is and diverges, then diverges.

The first case is the most important one, and fortunately it will work even if you accidentally write instead of . The second and third cases require that you get the fraction "right side up".

Example. Determine whether converges or diverges.

The series has positive terms.

When k is large, the top and bottom are dominated by the terms with the biggest powers:

Compute the limiting ratio:

The limiting ratio is 1, a finite positive number. The series

diverges, because it is harmonic. Hence, the series diverges by Limit Comparison.

Example. Determine whether converges or diverges.

The series has positive terms.

When k is large,

Compute the limiting ratio:

The limiting ratio is 1, a finite positive number. The series

is a convergent geometric series (since ). Therefore, converges, by Limit Comparison.

Example. Determine whether converges or diverges.

It is possible to use Limit Comparison, comparing the series to the convergent geometric series . After a long computation, you'll find the limiting ratio is , so the test works.

However, there is a better way, which I'll discuss next. It is called the Root Test, and you perform it by taking the root of the term, then taking the limit as . If the limit is less than 1, the series converges. In this case,

Therefore, the series converges, by the Root Test.

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