/*! 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 87 A glimpse ahead to power series.... [FREE SOLUTION] | 91Ó°ÊÓ

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A glimpse ahead to power series. Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{2^{k}}$$

Short Answer

Expert verified
Answer: The power series converges for \(0 \leq x < 2\).

Step by step solution

01

In this exercise, the general term of the series is given by: $$a_k = \frac{x^{k}}{2^{k}}$$. #Step 2: Determine the ratio#

Find the ratio \(\frac{a_{k+1}}{a_k}\). This can be calculated by dividing the expression for \(a_{k+1}\) by the expression for \(a_k\): $$\frac{a_{k+1}}{a_k}=\frac{\frac{x^{k+1}}{2^{k+1}}}{\frac{x^{k}}{2^{k}}}$$. #Step 3: Simplify the ratio#
02

Simplify the expression obtained in step 2: $$\frac{a_{k+1}}{a_k} = \frac{x^{k+1}}{2^{k+1}} \cdot \frac{2^k}{x^k} = \frac{x}{2}$$. #Step 4: Apply the Ratio Test#

For a given series to converge using the Ratio Test, we need to find the limit: $$\lim_{k\rightarrow \infty} \frac{a_{k+1}}{a_k}$$ and check if it is less than 1: $$\lim_{k\rightarrow \infty} \frac{x}{2} = \frac{x}{2}$$. Since the limit does not depend on \(k\), it is constant and equal to \(\frac{x}{2}\). #Step 5: Determine the converging values of x#
03

In order to satisfy the Ratio Test, the limit should be less than 1: $$\frac{x}{2} < 1$$. Solve for x: $$x < 2$$. #Step 6: Present the final answer#

The series converges for all values of x, with \(x \geq 0\) and \(x < 2\). Hence, the power series converges for \(0 \leq x < 2\).

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Key Concepts

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

Ratio Test
The Ratio Test is a helpful tool in determining if a power series converges. It is based on considering the ratio of successive terms in the series. For the given series, our goal is to examine the limit of \(\frac{a_{k+1}}{a_k}\). This involves taking the term \(a_k = \frac{x^{k}}{2^{k}}\) and finding its successive term \(a_{k+1}\). By taking the ratio \(\frac{a_{k+1}}{a_k}\) and simplifying it, we obtain \(\frac{x}{2}\). The Ratio Test tells us that the series will converge if the absolute value of this limit is less than 1. Thus, we solve \(\frac{x}{2} < 1\) to find the range of values for \(x\) where the series converges. This condition gives us \(x < 2\).

Using the Ratio Test, we find that every term in the solution relies on simplifying the ratio of successive terms to learn about the behavior of the entire series as it progresses.
Convergence
Convergence is the key concept when dealing with power series. When a series converges, the sum of its terms approaches a finite number as more terms are added.

In mathematical terms, a series \(\sum_{k=1}^{\infty} a_k\) converges if the sequence of its partial sums \(S_n = a_1 + a_2 + \ldots + a_n\) approaches a limit as \(n\) becomes very large. For the exercise in question, convergence is determined using our outcome from the Ratio Test, where we found that the terms become small enough as \(x\) approaches values less than 2.

For a series, knowing the range of \(x\) where it converges allows us to understand how the series behaves overall. This involves confirming that the sum remains finite under these conditions. In simple terms: convergence signifies a kind of balance or result that isn't infinite in scope.
Series Convergence
Series convergence involves analyzing the whole series and understanding when it behaves in a predictable manner. It's essential to know the interval of \(x\) for which the infinite sum \(\sum_{k=1}^{\infty} \frac{x^k}{2^k}\) remains manageable, meaning the series sum doesn't spiral out to infinity.

The Ratio Test gave us clarity by showing that when \(x < 2\), the series \(\sum_{k=1}^{\infty} \frac{x^k}{2^k}\) is manageable and converges. This finding is significant because it tells us precisely which values of \(x\) keep the series from diverging, or in simpler terms, from exploding to infinity.

Understanding series convergence isn't just about setting limits but grasping where these limits are beneficial and ensure the series sums to a finite value. Notably, we consider the interval \(0 \leq x < 2\) as the realm where this power series converges.
Limit of a Sequence
The concept of the limit of a sequence is foundational when talking about series and their convergence.

For a sequence \(a_k\), the limit is determined as \(k\) approaches infinity. This limit helps identify where the terms are heading and if they are ultimately converging to a specific number. In the context of the exercise, we examine \(\lim_{k \to \infty} \frac{x}{2}\), which simplifies to \(\frac{x}{2}\) since it isn’t dependent on \(k\), and implies that all terms trend toward becoming infinitesimally small under the right conditions of \(x \lt 2\). This calculation is essential for the Ratio Test and series convergence.

Understanding limits is about seeing the big picture in small components. For a series, each small part (or sequence term) helps us predict the broader behavior of the series sum and is crucial in ensuring convergence.

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Most popular questions from this chapter

James begins a savings plan in which he deposits \(\$ 100\) at the beginning of each month into an account that earns \(9 \%\) interest annually or, equivalently, \(0.75 \%\) per month. To be clear, on the first day of each month, the bank adds \(0.75 \%\) of the current balance as interest, and then James deposits \(\$ 100\). Let \(B_{n}\) be the balance in the account after the \(n\) th deposit, where \(B_{0}=\$ 0\). a. Write the first five terms of the sequence \(\left\\{B_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{B_{n}\right\\}\). c. How many months are needed to reach a balance of \(\$ 5000 ?\)

Consider the number \(0.555555 \ldots,\) which can be viewed as the series \(5 \sum_{k=1}^{\infty} 10^{-k} .\) Evaluate the geometric series to obtain a rational value of \(0.555555 .\) b. Consider the number \(0.54545454 \ldots\), which can be represented by the series \(54 \sum_{k=1}^{\infty} 10^{-2 k} .\) Evaluate the geometric series to obtain a rational value of the number. c. Now generalize parts (a) and (b). Suppose you are given a number with a decimal expansion that repeats in cycles of length \(p,\) say, \(n_{1}, n_{2} \ldots ., n_{p},\) where \(n_{1}, \ldots, n_{p}\) are integers between 0 and \(9 .\) Explain how to use geometric series to obtain a rational form for \(0 . \overline{n_{1}} n_{2} \cdots n_{p}\) d. Try the method of part (c) on the number \(0 . \overline{123456789}=0.123456789123456789 \ldots\) e. Prove that \(0 . \overline{9}=1\)

Evaluate the series \(\sum_{k=1}^{\infty}\left(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right)\) two ways. a. Use a telescoping series argument. b. Use a geometric series argument after first simplifying \(\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\)

Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \dots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots$$

Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\) (Exercises 65 and 66 ) and that the terms of this series may be rearranged without changing the value of the series. Determine the sum of the reciprocals of the squares of the odd positive integers.

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