/*! 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 66 Determine whether each infinite ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 66

Determine whether each infinite geometric series converges or diverges. If it converges, find its sum. $$ \sum_{k=1}^{\infty} 4\left(-\frac{1}{2}\right)^{k-1} $$

Short Answer

Expert verified
The series converges, and its sum is \( \frac{8}{3} \).

Step by step solution

01

Identify the General Form of the Series

The given series can be written in the general form of a geometric series: the sum of the series can be expressed as the first term, denoted as \(a\), and common ratio, denoted as \(r\).
02

Determine the First Term

Identify the first term (\( a \)) of the series: For the given series: \( 4\bigg(-\frac{1}{2}\bigg)^{k-1} \), the first term is the value of the series when \( k = 1 \): \( a = 4\bigg(-\frac{1}{2}\bigg)^{1-1} = 4 \).
03

Determine the Common Ratio

Identify the common ratio (\( r \)) of the series: The common ratio is the factor by which each term is multiplied to get the next term. For the given series: \( 4\bigg(-\frac{1}{2}\bigg)^{k-1} \): \( r = -\frac{1}{2} \).
04

Check for Convergence

A geometric series converges if the absolute value of the common ratio \( r \) is less than 1. Verify |\( r \)|: \( |-\frac{1}{2}| = \frac{1}{2} < 1 \), so the series converges.
05

Calculate the Sum

Use the formula for the sum of an infinite geometric series: \[ S = \frac{a}{1 - r} \]We know \( a = 4 \) and \( r = -\frac{1}{2} \). Substitute these values into the formula: \[ S = \frac{4}{1 - (-\frac{1}{2})} = \frac{4}{1 + \frac{1}{2}} = \frac{4}{\frac{3}{2}} = \frac{4 \times 2}{3} = \frac{8}{3} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

convergence of series
When studying infinite geometric series, one key concept is convergence. A series converges if it approaches a finite value as more and more terms are added. For a geometric series, convergence is determined by the common ratio, denoted as \( r \).
To check if a geometric series converges, examine the absolute value of the common ratio \( |r| \).
The series converges if and only if \( |r| < 1 \).
  • For example, in the series \( \sum_{k=1}^{\infty} 4\left(-\frac{1}{2}\right)^{k-1} \), the common ratio \( r \) is \( -\frac{1}{2} \).
  • Since \( | -\frac{1}{2} | = \frac{1}{2} \), which is less than 1, the series converges.
In contrast, if \( |r| \geq 1 \), the series diverges, meaning it does not approach a finite value.
sum of infinite series
The sum of an infinite geometric series can be found only if the series converges. For a convergent geometric series, there is a simple formula to calculate the sum.
This formula sums all the terms from the first to infinity, resulting in a finite value.
  • The general formula for the sum \( S \) of an infinite geometric series is \[ S = \frac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio.
Let's apply this to the series \( 4\left(-\frac{1}{2}\right)^{k-1} \):
  • First term \( a = 4 \)
  • Common ratio \( r = -\frac{1}{2} \)
Substituting into the formula, we get: \[ S = \frac{4}{1 - (-\frac{1}{2})} = \frac{4}{1 + \frac{1}{2}} = \frac{4}{\frac{3}{2}} = \frac{8}{3} \].
So, the sum of the series is \( \frac{8}{3} \).
geometric series formula
A geometric series has a distinct pattern where each term after the first is obtained by multiplying the previous term by a constant, known as the common ratio \( r \).

The general form of a geometric series is:
\sum_{k=1}^{\infty} ar^{k-1}, where:
  • \( a \) is the first term
  • \( r \) is the common ratio
For example, in the series \( 4\left(-\frac{1}{2}\right)^{k-1} \):
  • First term \( a = 4 \)
  • Common ratio \( r = -\frac{1}{2} \)
Each term in this series is obtained by multiplying the previous term by \( -\frac{1}{2} \).
Understanding this pattern is crucial for identifying and working with geometric series.
This formula allows us to find sums and analyze the behavior of series effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bode's Law In \(1772,\) Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun: $$ a_{1}=0.4 \quad a_{n}=0.4+0.3 \cdot 2^{n-2} $$ where \(n \geq 2\) is the number of the planet from the sun. (a) Determine the first eight terms of the sequence. (b) At the time of Bode's publication, the known planets were Mercury \((0.39 \mathrm{AU}),\) Venus \((0.72 \mathrm{AU}),\) Earth \((1 \mathrm{AU})\) Mars \((1.52 \mathrm{AU}),\) Jupiter \((5.20 \mathrm{AU}),\) and Saturn \((9.54 \mathrm{AU})\) How do the actual distances compare to the terms of the sequence? (c) The planet Uranus was discovered in \(1781,\) and the asteroid Ceres was discovered in \(1801 .\) The mean orbital distances from the sun to Uranus and Ceres " are \(19.2 \mathrm{AU}\) and \(2.77 \mathrm{AU},\) respectively. How well do these values fit within the sequence? (d) Determine the ninth and tenth terms of Bode's sequence. (e) The planets Neptune and Pluto" were discovered in 1846 and \(1930,\) respectively. Their mean orbital distances from the sun are \(30.07 \mathrm{AU}\) and \(39.44 \mathrm{AU},\) respectively. How do these actual distances compare to the terms of the sequence? (f) On July \(29,2005,\) NASA announced the discovery of a dwarf planet \((n=11),\) which has been named Eris. Use Bode's Law to predict the mean orbital distance of Eris from the sun. Its actual mean distance is not yet known, but Eris is currently about 97 astronomical units from the sun.

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find a rectangular equation of the plane curve with parametric equations \(x(t)=t+5\) and \(y(t)=\sqrt{t}\) for \(t \geq 0\).

Challenge Problem If the terms of a sequence have the property that \(\frac{a_{1}}{a_{2}}=\frac{a_{2}}{a_{3}}=\cdots=\frac{a_{n-1}}{a_{n}},\) show that \(\frac{a_{1}^{n}}{a_{2}^{n}}=\frac{a_{1}}{a_{n+1}}\)

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Use the Change-of-Base Formula and a calculator to evaluate \(\log _{7} 62\). Round the answer to three decimal places.

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ \frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\cdots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1} $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.