/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Determine whether the series con... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty} \frac{4^{n+1}}{3^{n}-2} $$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Identify the Series Type

The series is a sum of the form \( \sum_{n=1}^{\infty} \frac{4^{n+1}}{3^{n}-2} \), which suggests it is similar to a geometric series, but with an adjusted denominator. We can compare this to a simpler geometric series to analyze convergence.
02

Simplify the Expression

First, rewrite the fraction to see if it resembles a common test series. The expression \( \frac{4^{n+1}}{3^n - 2} \) can be compared to \( \frac{4^{n+1}}{3^n} \) for large \( n \), which simplifies to \( 4 \times \frac{4^n}{3^n} = 4 \left( \frac{4}{3} \right)^n \).
03

Determine Series Convergence using Ratio Test

Consider focusing on the simplified series \( \sum_{n=1}^{\infty} 4 \left( \frac{4}{3} \right)^n \). Apply the ratio test to this geometric series to check convergence. The ratio test states that if the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) is greater than 1, the series diverges.
04

Apply the Ratio Test

For the geometric component \( a_n = 4 \left( \frac{4}{3} \right)^n \), calculate \( \frac{a_{n+1}}{a_n} = \frac{4 \left( \frac{4}{3} \right)^{n+1}}{4 \left( \frac{4}{3} \right)^n} = \frac{4}{3} \). Clearly, \( \frac{4}{3} > 1 \), indicating divergence.
05

Conclusion about the Original Series

Since \( 4 \left( \frac{4}{3} \right)^n \) diverges, and \( \frac{4^{n+1}}{3^n - 2} \) for large \( n \) behaves like this series, the original series also diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Series
A geometric series is a series where each term, after the first, is found by multiplying the previous one by a fixed, non-zero number known as the common ratio. This type of series can be denoted as \( a, ar, ar^2, ar^3, \ldots \) where \( a \) is the first term and \( r \) is the common ratio. The convergence of a geometric series depends entirely on the value of \( r \):
  • If \( |r| < 1 \), the series converges.
  • If \( |r| \geq 1 \), the series diverges.
To use this for analysis, simply compare the given series to a geometric series by rearranging or simplifying its terms to highlight any geometric relationships. This will help determine if a convergence or divergence analysis is feasible. Understanding this fundamental concept is crucial for tackling more complex series, like the one given in the exercise.
Ratio Test
The ratio test is a powerful method used to determine the convergence or divergence of an infinite series. Given a series with terms \( a_n \), the ratio test involves evaluating the limit:\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]The results of the ratio test are:
  • If \( L < 1 \), the series converges.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive, and further analysis is needed.
In the step-by-step solution provided, the ratio test was applied to the simplified geometric-like series to show divergence. Calculating \( L = \frac{4}{3} \), which is greater than 1, confirmed that the series diverges. This procedure reveals the efficiency of the ratio test, especially when handling series closely resembling geometric ones.
Divergence of Series
The concept of divergence is crucial when analyzing series. A series is said to diverge if the sum of its terms increases without bound as more terms are added. This can occur when series terms even out or maintain significant non-negating contributions. In the context of the exercise, both the original and simplified series were assessed for divergence. Using the simplified expression and analysis, it was found that terms grow in a manner consistent with a divergent series, particularly since the ratio exceeded 1. Recognizing divergence is useful because it informs whether continued summation efforts will lead to a definitive sum or result in the series spiraling towards infinity. Being able to identify this helps mathematicians and students make informed decisions about the behaviors of series they encounter.
Limit Comparison
The limit comparison test is a handy tool for determining the convergence or divergence of more complex series that don’t fit neatly into geometric or arithmetic categories. It is employed by comparing the series of interest to another series whose convergence status is already known.To use the limit comparison test, you consider two series with positive terms, say \( \sum a_n \) and \( \sum b_n \). You calculate:\[ L = \lim_{n \to \infty} \frac{a_n}{b_n} \]Where:
  • If \( 0 < L < \infty \), then both series either converge or diverge together.
  • If \( L = 0 \) or \( L = \infty \), the test is inconclusive.
In the exercise, understanding the behavior of similar simpler series, like the geometric series involved, helps inform the decision about the original series' divergence. The limit comparison ensures a deeper and often more intuitive understanding by relating a complex series to a more straightforward, well-understood one.

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