Problem 86 - Essential Calculus

Question

State the Limit Comparison Test and give an example of its use.

Step-by-Step Solution

Answer

The Limit Comparison Test helps to determine whether a series converges or diverges by comparing it to another series. If the limit of the ratio of two series is a finite positive number, then both series either converge or diverge. Here, for the series \( \sum_{n=1}^{\infty}\frac{n}{n^2+1} \), we compared it with the series \( \frac{1}{n} \) and found that both series diverge.

Step 1: Definition of Limit Comparison Test

The Limit Comparison Test is a method in calculus used to determine whether a series converges or diverges. Let \( a_n \) and \( b_n \) be two series where \( b_n > 0 \). Calculate the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \). If \( L \) is a finite positive number (i.e., \( 0 < L < \infty \)), then both series either converge or diverge.

Step 2: Example of Limit Comparison Test

Let's illustrate this with an example. Consider the series \( \sum_{n=1}^{\infty}\frac{n}{n^2+1} \) . To use the Limit Comparison Test, we need to select a second series. A good choice is \( b_n = \frac{1}{n} \), which is known to diverge. We calculate the limit: \( L = \lim_{n \to \infty} \frac{(n/(n^2+1))}{(1/n)} = \lim_{n \to \infty} \frac{n^2}{n^2 + 1} = 1 \), which is a finite positive number. Thus, our original series behaves in the same way as \( \frac{1}{n} \), which is to say it diverges.