Problem 50 - Essential Calculus

Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{4^{n}}{3^{n}-1} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{4^{n}}{3^{n}-1} \) diverges by the Direct Comparison Test.

Step 1: Simplify the Series

First, start by simplifying the series we are given: \( \sum_{n=1}^{\infty} \frac{4^{n}}{3^{n}-1} \). Since \( 3^{n}-1 \) is less than \( 3^{n} \), our series is larger than the series \( \sum_{n=1}^\infty \frac{4^{n}}{3^n} \).

Step 2: Apply the Direct Comparison Test

In the Direct Comparison Test, if we have a series \( a_n \) that we know diverges and a second series \( b_n \) such that \( 0 ≤ b_n ≤ a_n \), then \( b_n \) also diverges. Here, our series \( \frac{4^{n}}{3^{n}-1} \) corresponds to \( b_n \) and \( \frac{4^{n}}{3^{n}} \) corresponds to \( a_n \).

Step 3: Compare the series

We know by the geometric series test that the series \( \frac{4^{n}}{3^{n}} \) diverges because \( \frac{4}{3} \) is greater than 1. Therefore, since the series \( \frac{4^{n}}{3^{n}-1} \) is larger than a series known to diverge, \( \frac{4^{n}}{3^{n}-1} \), by the Direct Comparison Test, it also diverges.