# What is the Limit Comparison Test?

Jul 14, 2015

It's a test, involving limits of sequences, used to decide whether a given series $\setminus {\sum}_{n = 1}^{\setminus \infty} {a}_{n}$ converges or not, based on knowledge that another (often related) series $\setminus {\sum}_{n = 1}^{\setminus \infty} {b}_{n}$ converges.

#### Explanation:

Here's a full statement of the test:

Suppose $\setminus {\sum}_{n = 1}^{\setminus \infty} {a}_{n}$ and $\setminus {\sum}_{n = 1}^{\setminus \infty} {b}_{n}$ are two series with positive terms (that is, ${a}_{n} > 0$ and ${b}_{n} > 0$ for all $n$). Let ${r}_{n} = {a}_{n} / {b}_{n}$. Then:

a) If $\setminus {\sum}_{n = 1}^{\setminus \infty} {b}_{n}$ converges and ${\lim}_{n \to \setminus \infty} {r}_{n}$ exists, then $\setminus {\sum}_{n = 1}^{\setminus \infty} {a}_{n}$ converges.

b) If $\setminus {\sum}_{n = 1}^{\setminus \infty} {b}_{n}$ diverges and ${\lim}_{n \to \setminus \infty} {r}_{n} > 0$ or $\setminus {\lim}_{n \to \setminus \infty} {r}_{n} = \setminus \infty$, then $\setminus {\sum}_{n = 1}^{\setminus \infty} {a}_{n}$ diverges.

As an simple example, suppose you wish to know whether the series $\setminus {\sum}_{n = 1}^{\setminus \infty} \frac{5}{2 {n}^{2} - 1}$ converges or not. This series is somewhat similar to the p-series $\setminus {\sum}_{n = 1}^{\setminus \infty} \frac{1}{n} ^ 2$, which is known to converge. Let ${a}_{n} = \frac{5}{2 {n}^{2} - 1}$ and ${b}_{n} = \frac{1}{n} ^ 2$ so that ${r}_{n} = \frac{5 {n}^{2}}{2 {n}^{2} - 1}$. Since ${r}_{n} \to \frac{5}{2}$ as $n \to \setminus \infty$, it follows from the limit comparison test that $\setminus {\sum}_{n = 1}^{\setminus \infty} \frac{5}{2 {n}^{2} - 1}$ converges as well. If you are familiar with the "regular" comparison test, note that the limit comparison test is a bit simpler to use for this example.