/*! 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 41 Decide which of the following ge... [FREE SOLUTION] | 91Ó°ÊÓ

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Decide which of the following geometric series are convergent. (a) \(2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\ldots+\frac{2}{3^{k}}+\ldots\) (b) \(4-2+1-\frac{1}{2}+\ldots+\frac{(-1)^{k} 4}{2^{k}}+\ldots\) (c) \(10+11+\frac{121}{10}+\frac{1331}{100}+\ldots+10\left(\frac{11}{10}\right)^{k}+\ldots\) (d) \(1-\frac{5}{4}+\frac{25}{16}-\frac{125}{64}+\ldots+\left(\frac{-5}{4}\right)^{k}+\ldots\)

Short Answer

Expert verified
(a) and (b) are convergent; (c) and (d) are divergent.

Step by step solution

01

Identifying the Series

Each series can be written in the form \( a + ar + ar^2 + ar^3 + \ldots \), which is a geometric series. We need to determine the first term \( a \) and the common ratio \( r \) for each series given in the problem.
02

Analyzing Series (a)

The series (a) is given as \( 2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\ldots+\frac{2}{3^{k}}+\ldots\), where the first term is \( a = 2 \) and the common ratio \( r = \frac{1}{3} \). A geometric series converges if \(|r| < 1\), which in this case \(|\frac{1}{3}| = \frac{1}{3} < 1\). Hence, this series is convergent.
03

Analyzing Series (b)

The series (b) is given as \( 4-2+1-\frac{1}{2}+\ldots+\frac{(-1)^{k} 4}{2^{k}}+\ldots\), meaning \( a = 4 \) and \( r = -\frac{1}{2} \). Since \(|r| = \frac{1}{2} < 1\), this series is convergent.
04

Analyzing Series (c)

The series (c) is \( 10+11+\frac{121}{10}+\frac{1331}{100}+\ldots+10\left(\frac{11}{10}\right)^{k}+\ldots\), where \( a = 10 \) and \( r = \frac{11}{10} \). Since \(|r| = \frac{11}{10} > 1\), this series is divergent.
05

Analyzing Series (d)

The series (d) is \( 1-\frac{5}{4}+\frac{25}{16}-\frac{125}{64}+\ldots+\left(\frac{-5}{4}\right)^{k}+\ldots\), with \( a = 1 \) and \( r = -\frac{5}{4} \). Since \(|r| = \frac{5}{4} > 1\), this series is divergent.

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Key Concepts

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

Convergence of Series
Understanding when a geometric series converges is crucial for analyzing infinite series. A geometric series can be written 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 on the absolute value of the common ratio. To determine if a series converges, we check the condition \( |r| < 1 \). If this condition is met, the series is convergent, meaning the sum approaches a specific finite value as the number of terms goes to infinity.
Let's use some examples to illustrate this principle. In the first series from our exercise, \( 2+\frac{2}{3}+\frac{2}{9}+\ldots \), the common ratio is \( \frac{1}{3} \), which satisfies \( |\frac{1}{3}| < 1 \). Therefore, this first series is convergent. Another example is the second series, \( 4-2+1-\frac{1}{2}+\ldots \), with a common ratio \( -\frac{1}{2} \). Despite being negative, because \( |-\frac{1}{2}| = \frac{1}{2} < 1 \), this series is also convergent.
To sum up, for any series to converge, always examine the absolute value of the common ratio. If less than one, convergence is guaranteed.
Common Ratio
In geometric series, the common ratio \( r \) is the factor by which each term in the series is multiplied to get the subsequent term. This ratio is an essential element that dictates the behavior of the series.

To identify the common ratio, examine any term of the series divided by the previous term. For instance, in the series \( 2+\frac{2}{3}+\frac{2}{9}+\ldots \), the second term divided by the first term gives the common ratio \( \frac{1}{3} \). Similarly, the sequence \( 4-2+1-\frac{1}{2}+\ldots \) has a common ratio of \( -\frac{1}{2} \) seen by dividing the second term \(-2\) by the first term \(4\).
  • If \( |r| < 1 \), the series converges.
  • If \( |r| = 1 \), the series does not converge since it does not settle at a sum.
  • If \( |r| > 1 \), the series diverges.
Identifying and analyzing the common ratio allows us to easily determine whether a series will converge or diverge.
Divergence of Series
Understanding when a series diverges is just as important as understanding convergence. Divergence occurs when the sum of the series does not approach any finite number as more terms are added.

In geometric series, divergence is determined by the common ratio \( r \). Specifically, a series diverges if the absolute value of the common ratio is greater than one, \( |r| > 1 \). An example from our exercise is the series \( 10+11+\frac{121}{10}+\frac{1331}{100}+\ldots \), where \( r = \frac{11}{10} \). The absolute value of this ratio is greater than one, causing the series to diverge. Furthermore, the series \( 1-\frac{5}{4}+\frac{25}{16}-\frac{125}{64}+\ldots \), with \( r = -\frac{5}{4} \), also diverges since \( |\frac{5}{4}| > 1 \).

Diverging series lack a finite sum and essentially run off to infinity (if they're positive) or accumulate to an undefined sum. This characteristic makes it crucial to analyze the common ratio, as it determines the very nature of the series' behavior.

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Most popular questions from this chapter

A sequence is defined by \(x_{n}=p n+q\) where \(p\) and \(q\) are constants. If \(x_{2}=7\) and \(x_{8}=-11\), find \(p\) and \(q\) and write down (a) the first four terms of the sequence; (b) the defining recurrence relation for the sequence.

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