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In Exercises \(1-8,\) use the Ratio Test to determine if each series conyerges ahsolutely or diveroes. $$\sum_{n=1}^{\infty} \frac{2^{n+1}}{n 3^{n-1}}$$

Short Answer

Expert verified
The series converges absolutely by the Ratio Test.

Step by step solution

01

Identify the Terms of the Series

The series given is \( \sum_{n=1}^{\infty} \frac{2^{n+1}}{n 3^{n-1}} \). We observe that the general term \( a_n \) can be written as \( a_n = \frac{2^{n+1}}{n 3^{n-1}} \).
02

Simplify the General Term

We simplify \( a_n \) as follows:\[a_n = \frac{2^{n+1}}{n 3^{n-1}} = \frac{2 \cdot 2^{n}}{n \cdot \frac{3^{n}}{3}} = \frac{6 \cdot 2^n}{n \cdot 3^n} = \frac{6 \left(\frac{2}{3}\right)^n}{n}.\]
03

Apply the Ratio Test

The Ratio Test involves computing the limit \( L \) of the absolute value of the ratio of successive terms. Compute:\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{6 \left(\frac{2}{3}\right)^{n+1}}{n+1} \times \frac{n}{6 \left(\frac{2}{3}\right)^n} \right|.\]
04

Simplify the Ratio

Cancel out the terms in the fraction and simplify:\[L = \lim_{n \to \infty} \frac{n}{n+1} \left(\frac{2}{3}\right) = \lim_{n \to \infty} \frac{n}{n+1} \cdot \frac{2}{3}.\]This simplifies further to:\[L = \frac{2}{3} \lim_{n \to \infty} \frac{n}{n+1} = \frac{2}{3} \times 1 = \frac{2}{3}.\]
05

Conclude with the Ratio Test

According to the Ratio Test, if \( L < 1 \), then the series converges absolutely. Here, since \( L = \frac{2}{3} \), which is less than 1, the series \( \sum_{n=1}^{\infty} \frac{2^{n+1}}{n 3^{n-1}} \) converges absolutely.

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Key Concepts

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

Absolute Convergence
Absolute convergence is a central topic in the study of series. When we say a series converges absolutely, it means that not only does the series converge, but it also converges if we take the absolute value of every term.

This is a stronger statement than simple convergence. Consider a series \( \sum_{n=1}^{\infty} a_n \). If \( \sum_{n=1}^{\infty} |a_n| \) converges, then the series \( \sum_{n=1}^{\infty} a_n \) is said to converge absolutely.

Why is this important? Because if a series converges absolutely, it means the series will also converge even if we change the order of its terms. This property is quite useful for proving the convergence of series related to complex numbers or when dealing with series that have both positive and negative terms.
  • Absolute convergence guarantees convergence irrespective of term arrangement.
  • It is indicative of a series’ stability under term rearrangement.
Infinite Series
Infinite series are sums with infinitely many terms, generally in the form \( \sum_{n=1}^{\infty} a_n \). Their analysis seeks to understand if these series converge to a finite limit, diverge, or do something else altogether.

Understanding the convergence or divergence of infinite series is crucial, as it determines whether the sum has a meaningful finite value. Various tests can be applied to determine convergence, including the Ratio Test used in the provided solution.
  • Infinite series can converge to a limit or diverge.
  • They are extensively studied using various tests.
  • Analysis of these series is essential for calculus and other advanced mathematics fields.
Being able to determine convergence swiftly with tools like the Ratio Test is important for dealing with complex series, particularly where terms grow or shrink based on an exponential factor, as you can see in the problem given.
Series Convergence
Series convergence refers to the behavior of the sum \( \sum_{n=1}^{\infty} a_n \) as the number of terms approaches infinity. If the sum approaches a specific number (finite limit), the series is said to converge.

The Ratio Test, used here, is a powerful tool for testing series convergence, particularly when the series terms include exponential components.

The Ratio Test Steps:

1. Identify the general term \( a_n \) of the series.2. Compute the limit of \( \left| \frac{a_{n+1}}{a_n} \right| \) as \( n \to \infty \).3. Analyze the resulting limit \( L \):
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \) or \( L \) is infinite, the series diverges.
  • If \( L = 1 \), the test is inconclusive.
In this exercise, after simplifying the ratio of successive terms, the limit \( L \) was found to be \( \frac{2}{3} \), indicating absolute convergence.
  • Convergence testing helps establish whether a series can be summed to a finite value.
  • Understanding each test is key for determining the behavior of the series.

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Most popular questions from this chapter

How many terms of the convergent series \(\sum_{n=1}^{\infty}\left(1 / n^{1.1}\right)\) should be used to estimate its value with error at most 0.00001\(?\)

Logarithmic \(p\) -series a. Show that the improper integral $$\int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant })$$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.

According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan \(^{-1} 1\) would you have to add to be sure of finding \(\pi / 4\) with an error of magnitude less than \(10^{-3} ?\) Give reasons for your answer.

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{2^{n}-n}{n 2^{n}}\end{equation}

Computer Explorations Taylor's formula with \(n=1\) and \(a=0\) gives the linearization of a function at \(x=0 .\) With \(n=2\) and \(n=3\) we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions: \begin{equation} \begin{array}{l}{\text { a. For what values of } x \text { can the function be replaced by each }} \\ {\text { approximation with an error less than } 10^{-2} \text { ? }} \\ {\text { b. What is the maximum error we could expect if we replace the }} \\ {\text { function by each approximation over the specified interval? }}\end{array} \end{equation} Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals in Exercises \(53-58 .\) \begin{equation} \begin{array}{l}{\text { Step } 1 : \text { Plot the function over the specified interval. }} \\ {\text { Step } 2 : \text { Find the Taylor polynomials } P_{1}(x), P_{2}(x), \text { and } P_{3}(x) \text { at }} \\\ {x=0 .}\\\\{\text { Step } 3 : \text { Calculate the }(n+1) \text { st derivative } f^{(n+1)}(c) \text { associated }} \\ {\text { with the remainder term for each Taylor polynomial. }} \\ {\text { Plot the derivative as a function of } c \text { over the specified interval }} \\ {\text { and estimate its maximum absolute value, } M .}\\\\{\text { Step } 4 : \text { Calculate the remainder } R_{n}(x) \text { for each polynomial. }} \\ {\text { Using the estimate } M \text { from Step } 3 \text { in place of } f^{(n+1)}(c), \text { plot }} \\ {R_{n}(x) \text { over the specified interval. Then estimate the values of }} \\ {x \text { that answer question (a). }}\\\\{\text { Step } 5 : \text { Compare your estimated error with the actual error }} \\ {E_{n}(x)=\left|f(x)-P_{n}(x)\right| \text { by plotting } E_{n}(x) \text { over the specified }} \\ {\text { interval. This will help answer question (b). }} \\ {\text { Step } 6 : \text { Graph the function and its three Taylor approximations }} \\ {\text { together. Discuss the graphs in relation to the information }} \\ {\text { discovered in Steps } 4 \text { and } 5 .}\end{array} \end{equation} $$f(x)=(1+x)^{3 / 2},-\frac{1}{2} \leq x \leq 2$$

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