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Chapter 7: Problem 50

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

Short Answer

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

Step by step solution

01

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 \( \sum_{n=1}^\infty \frac{4^{n}}{3^n} \).
02

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 \( 0 鈮� b_n 鈮� a_n \), 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 \( a_n \).
03

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.

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Most popular questions from this chapter

Give an example of a sequence satisfying the condition or explain why no such sequence exists. (Examples are not unique.) A monotonically increasing bounded sequence that does not converge

State the \(n\) th-Term Test for Divergence.

Find the sum of the convergent series. $$ 1+0.1+0.01+0.001+\cdots $$

Compute the first six terms of the sequence \(\left\\{a_{n}\right\\}=\\{\sqrt[n]{n}\\} .\) If the sequence converges, find its limit.

State the definitions of convergent and divergent series.
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