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 $\displaystyle \sum_{k=1}^\infty a_k$ , $\displaystyle \sum_{k=1}^\infty b_k$ , be series with positive terms.

  1. If $a_k \ge b_k$ for all k and $\displaystyle \sum_{k=1}^\infty a_k$ converges, then $\displaystyle \sum_{k=1}^\infty b_k$ converges.
  1. If $a_k \ge b_k$ for all k and $\displaystyle \sum_{k=1}^\infty b_k$ diverges, then $\displaystyle \sum_{k=1}^\infty b_k$ diverges.

It's easy to see that these tests make sense. In the first case, the partial sums of $\displaystyle \sum_{k=1}^\infty b_k$ are bounded above by $\displaystyle \sum_{k=1}^\infty a_k$ . The $b_k$ partial sums increase, so they must converge.

In the second case, the $a_k$ partial sums are always bigger than the $b_k$ partial sums, but the $b_k$ partial sums go to $\infty$ . Hence, the $a_k$ partial sums go to $\infty$ as well.

Example. Determine whether $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^3 + k + 7}$ converges or diverges.

The series has positive terms. In fact, I could use the Integral Test, but who would want to integrate $\dfrac{1}{x^3 + x + 7}$ ?

Instead, note that when k is large, the $k^3$ term should dominate. How does $\dfrac{1}{k^3 + k + 7}$ compare to $\dfrac{1}{k^3}$ ? Well, if you make the bottom smaller, the fraction gets bigger:

$$\dfrac{1}{k^3 + k + 7} < \dfrac{1}{k^3}.$$

Now $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^3}$ is a p-series with $p = 3$ , so it converges. Hence, $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^3 + k + 7}$ converges by comparison.

Example. Determine whether $\displaystyle \sum_{k=17}^\infty \dfrac{(\sin k)^2}{k(k + 1)}$ converges or diverges.

The series has positive terms. Since $(\sin k)^2 \le 1$ ,

$$\dfrac{(\sin k)^2}{k(k + 1)} \le \dfrac{1}{k(k + 1)} \le \dfrac{1}{k^2}.$$

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

Now $\displaystyle \sum_{k=17}^\infty \dfrac{1}{k^2}$ is a p-series with $p = 2$ , so it converges. Hence, $\displaystyle \sum_{k=17}^\infty \dfrac{(\sin k)^2}{k(k + 1)}$ converges by comparison.

Example. Determine whether $\displaystyle \sum_{k=1}^\infty \dfrac{\arctan k}{k^3}$ converges or diverges.

The series has positive terms. Since $\arctan k \le \dfrac{\pi}{2}$ ,

$$\dfrac{\arctan k}{k^3} \le \dfrac{1}{k^3}.$$

$\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^3}$ converges, because it's a p-series with $p = 3 > 1$ . Therefore, $\displaystyle \sum_{k=1}^\infty \dfrac{\arctan k}{k^3}$ converges by direct comparison.

Example. Determine whether $\displaystyle \sum_{k=1}^\infty \dfrac{\sqrt{k + 3}}{k\sqrt{k + 2}}$ converges or diverges.

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

$$\dfrac{\sqrt{k + 3}}{k\sqrt{k + 2}} > \dfrac{\sqrt{k + 2}}{k\sqrt{k + 2}} = \dfrac{1}{k}.$$

Notice how I avoided changing $k + 3$ to k; I changed it to something which cancelled the radical on the bottom.

Now $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k}$ diverges --- it's harmonic! So $\displaystyle \sum_{k=1}^\infty \dfrac{\sqrt{k + 3}}{k\sqrt{k + 2}}$ diverges, by comparison.

Example. Comparison won't work if the inequalities go the wrong way. For example, consider $\displaystyle \sum_{k=2}^\infty \dfrac{1}{k^2 - 2}$ . I'd like to compare this to $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^2}$ , but if I make the bottom bigger (by adding 2), the fraction gets smaller:

$$\dfrac{1}{k^2 - 2} > \dfrac{1}{k^2}.$$

It's true that $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^2}$ is a convergent p-series ($p = 2$ ), but it's smaller than the given series. I can't draw a conclusion this way.

Nevertheless, $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^2}$ 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 $\displaystyle \sum_{k=1}^\infty a_k$ be a positive term series.

Let $\displaystyle \sum_{k=1}^\infty b_k$ 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

$$\lim_{k \to \infty} \dfrac{a_k}{b_k}.$$

  1. If the limit is a finite positive number, then the two series behave in the same way. That is, if $\displaystyle \sum_{k=1}^\infty b_k$ converges, then $\displaystyle \sum_{k=1}^\infty a_k$ converges; if $\displaystyle \sum_{k=1}^\infty b_k$ diverges, then $\displaystyle \sum_{k=1}^\infty a_k$ diverges.
  2. If the limit is 0 and $\displaystyle \sum_{k=1}^\infty b_k$ converges, then $\displaystyle \sum_{k=1}^\infty a_k$ converges.
  3. If the limit is $+\infty$ and $\displaystyle \sum_{k=1}^\infty b_k$ diverges, then $\displaystyle \sum_{k=1}^\infty a_k$ diverges.

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

Example. Determine whether $\displaystyle \sum_{k=1}^\infty \dfrac{4k^2 + k + 9}{7k^3 + 13}$ 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:

$$\dfrac{4k^2 + k + 9}{7k^3 + 13} \approx \dfrac{4k^2}{7k^3} = \dfrac{4}{7}\cdot \dfrac{1}{k}.$$

Compute the limiting ratio:

$$\lim_{k \to \infty} \dfrac{\dfrac{4k^2 + k + 9} {7k^3 + 13}}{\dfrac{4}{7}\cdot \dfrac{1}{k}} = \dfrac{7}{4}\cdot \lim_{k \to \infty} \dfrac{4k^3 + k^2 + 9k}{7k^3 + 13} = \dfrac{7}{4}\cdot \dfrac{4}{7} = 1.$$

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

$$\sum_{k=1}^\infty \dfrac{4}{7}\cdot \dfrac{1}{k} = \dfrac{4}{7}\cdot \sum_{k=1}^\infty \dfrac{1}{k}$$

diverges, because it is harmonic. Hence, the series $\displaystyle \sum_{k=1}^\infty \dfrac{4k^2 + k + 9}{7k^3 + 13}$ diverges by Limit Comparison.

Example. Determine whether $\displaystyle \sum_{k=2}^\infty \dfrac{4^k + 5}{7^k - 42}$ converges or diverges.

The series has positive terms.

When k is large,

$$\dfrac{4^k + 5}{7^k - 42} \approx \dfrac{4^k}{7^k}.$$

Compute the limiting ratio:

$$\lim_{k \to \infty} \dfrac{\dfrac{4^k + 5}{7^k - 42}}{\dfrac{4^k}{7^k}} = \lim_{k \to \infty} \dfrac{1 + \dfrac{5}{4^k}}{1 - \dfrac{42}{7^k}} = 1.$$

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

$$\sum_{k=1}^\infty \dfrac{4^k}{7^k} = \sum_{k=1}^\infty \left(\dfrac{4}{7}\right)^k$$

is a convergent geometric series (since $\dfrac{4}{7} < 1$ ). Therefore, $\displaystyle \sum_{k=2}^\infty \dfrac{4^k + 5}{7^k - 42}$ converges, by Limit Comparison.

Example. Determine whether $\displaystyle \sum_{k=2}^\infty \left(\dfrac{\tan^{-1} k}{2}\right)^k$ converges or diverges.

It is possible to use Limit Comparison, comparing the series to the convergent geometric series $\displaystyle \sum_{k=2}^\infty \left(\dfrac{\pi}{4}\right)^k$ . After a long computation, you'll find the limiting ratio is $e^{-2/\pi}$ , 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 $k^{\rm th}$ root of the $k^{\rm th}$ term, then taking the limit as $k\to \infty$ . If the limit is less than 1, the series converges. In this case,

$$\lim_{k \to \infty} \root k \of {\left(\dfrac{\tan^{-1} k}{2}\right)^k} = \lim_{k \to \infty} \dfrac{\tan^{-1} k}{2} = \dfrac{\pi}{4} < 1.$$

Therefore, the series converges, by the Root Test.

Send comments about this page to: bikenaga@marauder.millersville.edu.

Last updated: December 2, 2005

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