/*! 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 1 Given: \(A_{n}=\frac{80}{8^{n}... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 1

Given: \(A_{n}=\frac{80}{8^{n}}\) Determine: (a) whether \(\sum_{n=1}^{\infty}\left(A_{n}\right)\) is convergent. (b) whether \(\left\\{A_{n}\right\\}\) is convergent. If convergent, enter the limit of convergence. If not, enter DIV.

Short Answer

Expert verified
The series converges. The sequence also converges to 0.

Step by step solution

01

Understanding the Sequence

Given: \(A_{n}=\frac{80}{8^{n}}\). Determine if the sequence \( \{A_{n}\} \) converges and if the series \( \sum_{n=1}^{\text{\textbackslash infinity}}\big(A_{n}\big) \) converges.
02

Rewrite the Sequence

Rewrite \(A_{n}\) as follows:\[ A_{n} = \frac{80}{8^{n}} = 80 \times \frac{1}{8^{n}} = 80 \times \frac{1}{(2^{3})^n} = 80 \times 2^{-3n} \]This form will help analyze the convergence of both the sequence and series.
03

Determine Sequence Convergence

For the sequence \( {A_{n}} \):A sequence \( {A_{n}} \) converges if its terms approach a specific value as \( n \) approaches infinity.Consider the behavior of \( A_{n} = 80 \times 2^{-3n} \) as \( n \to \infty \):As \( n \) increases, \( 2^{-3n} \) gets smaller and approaches 0.Thus, \( 80 \times 2^{-3n} \rightarrow 0 \).Therefore, \( \{A_{n}\} \) converges to 0.
04

Determine Series Convergence

To determine if the series \( \sum_{n=1}^{\text{\textbackslash infinity}}\big(A_{n}\big) \) converges, recognize that it is a geometric series:\[ A_{n} = \frac{80}{8^{n}} \Rightarrow \sum_{n=1}^{\text{\textbackslash infinity}}\big(A_{n}\big) = 80 \sum_{n=1}^{\text{\textbackslash infinity}}(2^{-3n}) \]A geometric series \( \sum_{n=1}^{\text{\textbackslash infinity}} ar^{n-1} \) converges if \( \left|r\right| < 1 \).Here, \( r = 2^{-3} = \left(\frac{1}{8}\right) \) which is less than 1.Thus, \( \sum_{n=1}^{\text{\textbackslash infinity}}\big(A_{n}\big) \) converges by the geometric series test.

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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 of terms where each term is a fixed multiple of the previous term. In general, a geometric series can be written as: . . . . In our example, the geometric series is: . . .
sequence convergence
Sequence convergence describes whether the terms of a sequence approach a particular value as the sequence progresses. For the sequence given in the exercise, we can observe its behavior by analyzing its formula: . . .
geometric series test
This test is essential to determine if an infinite geometric series converges or diverges. The geometric series test states that for a geometric series: will converge if the absolute value of the common ratio .
infinite series
An infinite series is a sum of infinite terms. Determining whether such a series converges (approaches a finite limit) or diverges (grows without bound) is critical in various areas of mathematics. Our example deals with an infinite series given by: . . .

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Most popular questions from this chapter

Airy's equation \(^{2}\) $$ y^{\prime \prime}-x y=0 $$ can be used to model an undamped vibrating spring with spring constant \(x\) (note that \(y\) is an unknown function of \(x\) ). So the solution to this differential equation will tell us the behavior of a spring-mass system as the spring ages (like an automobile shock absorber). Assume that a solution \(y=f(x)\) has a Taylor series that can be written in the form $$ y=\sum_{k=0}^{\infty} a_{k} x^{k} $$ where the coefficients are undetermined. Our job is to find the coefficients. (a) Differentiate the series for \(y\) term by term to find the series for \(y^{\prime}\). Then repeat to find the series for \(y^{\prime \prime}\). (b) Substitute your results from part (a) into the Airy equation and show that we can write Equation \((8.6 .4)\) in the form$$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}-\sum_{k=0}^{\infty} a_{k} x^{k+1}=0 $$ (c) At this point, it would be convenient if we could combine the series on the left in \((8.6 .5)\), but one written with terms of the form \(x^{k-2}\) and the other with terms in the form \(x^{k+1}\). Explain why $$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}=\sum_{k=0}^{\infty}(k+1)(k+2) a_{k+2} x^{k} $$ (d) Now show that $$ \sum_{k=0}^{\infty} a_{k} x^{k+1}=\sum_{k=1}^{\infty} a_{k-1} x^{k} $$ (e) We can now substitute \((8.6 .6)\) and \((8.6 .7)\) into \((8.6 .5)\) to obtain $$ \sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^{n}-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0 $$ Combine the like powers of \(x\) in the two series to show that our solution must satisfy $$ 2 a_{2}+\sum_{k=1}^{\infty}\left[(k+1)(k+2) a_{k+2}-a_{k-1}\right] x^{k}=0 $$ (f) Use equation (8.6.9) to show the following: i. \(a_{3 k+2}=0\) for every positive integer \(k\), ii. \(a_{3 k}=\frac{1}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)} a_{0}\) for \(k \geq 1\), iii. \(a_{3 k+1}=\frac{1}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)} a_{1}\) for \(k \geq 1\). (g) Use the previous part to conclude that the general solution to the Airy equation \((8.6 .4)\) is $$ \begin{aligned} y=& a_{0}\left(1+\sum_{k=1}^{\infty} \frac{x^{3 k}}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)}\right) \\ &+a_{1}\left(x+\sum_{k=1}^{\infty} \frac{x^{3 k+1}}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)}\right) \end{aligned} $$ Any values for \(a_{0}\) and \(a_{1}\) then determine a specific solution that we can approximate as closely as we like using this series solution.

Suppose you play a game with a friend that involves rolling a standard six- sided die. Before a player can participate in the game, he or she must roll a six with the die. Assume that you roll first and that you and your friend take alternate rolls. In this exercise we will determine the probability that you roll the first six. a. Explain why the probability of rolling a six on any single roll (including your first turn) is \(\frac{1}{6}\) b. If you don't roll a six on your first turn, then in order for you to roll the first six on your second turn, both you and your friend had to fail to roll a six on your first turns, and then you had to succeed in rolling a six on your second turn. Explain why the probability of this event is $$ \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right) $$ c. Now suppose you fail to roll the first six on your second turn. Explain why the probability is $$ \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right) $$ that you to roll the first six on your third turn. d. The probability of you rolling the first six is the probability that you roll the first six on your first turn plus the probability that you roll the first six on your second turn plus the probability that your roll the first six on your third turn, and so on. Explain why this probability is $$ \frac{1}{6}+\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right)+\cdots $$ Find the sum of this series and determine the probability that you roll the first six.

Compute the value of the following improper integral. If it converges, enter its value. Enter infinity if it diverges to \(\infty\), and -infinity if it diverges to \(-\infty\). Otherwise, enter diverges. $$ \int_{1}^{\infty} \frac{3 d x}{x^{2}+1}= $$ Does the series \(\sum_{n=1}^{\infty} \frac{3}{n^{2}+1}\) converge or diverge? [Choose: converges | diverges to +infinity | diverges to -infinity | diverges]

Find the first four terms of the Taylor series for the function \(\cos (x)\) about the point \(a=\) \(-\pi / 4\). (Your answers should include the variable \(\mathrm{x}\) when appropriate.) \(\cos (x)=\) \(+\ldots\)

Let $$ a_{n}=\frac{2 n}{10 n+7} $$ For the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter 'infinity' if it diverges to \(\infty\), '-infinity' if it diverges to \(-\infty\) or 'DNE' otherwise. (a) The series \(\sum_{n=1}^{\infty} \frac{2 n}{10 n+7}\). (b) The sequence \(\left\\{\frac{2 n}{10 n+7}\right\\}\).
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