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Converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$ \sum_{n=0}^{\infty} \frac{2^{n}-1}{3^{n}} $$

Short Answer

Expert verified
The series converges and its sum is \( \frac{3}{2} \).

Step by step solution

01

Identify the series form

The given series is \( \sum_{n=0}^{\infty} \frac{2^n - 1}{3^n} \). This series does not fit standard forms directly, so we should consider breaking it into separate parts that can be evaluated individually.
02

Split the series

Rewrite the series as two separate series using linearity of summation:\[ \sum_{n=0}^{\infty} \left( \frac{2^n}{3^n} - \frac{1}{3^n} \right) = \sum_{n=0}^{\infty} \left( \frac{2}{3} \right)^n - \sum_{n=0}^{\infty} \left( \frac{1}{3} \right)^n. \]
03

Analyze the first series

The first series is geometric with ratio \( r = \frac{2}{3} \). A geometric series \( \sum_{n=0}^\infty ar^n \) converges if \( |r| < 1 \). Since \( \left| \frac{2}{3} \right| < 1 \), this series converges. The sum is given by \[ \frac{1}{1 - \frac{2}{3}} = 3. \]
04

Analyze the second series

The second series is also geometric with ratio \( r = \frac{1}{3} \). Again, it converges since \( \left| \frac{1}{3} \right| < 1 \). The sum is given by \[ \frac{1}{1 - \frac{1}{3}} = \frac{3}{2}. \]
05

Calculate the total sum of the original series

Since both component series converge, we can find the total sum by subtracting their sums:\[ 3 - \frac{3}{2} = \frac{6}{2} - \frac{3}{2} = \frac{3}{2}. \]
06

Conclusion on convergence

Since both parts of the split series converge, the original series \( \sum_{n=0}^{\infty} \frac{2^n - 1}{3^n} \) converges as well, and its sum is \( \frac{3}{2} \).

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Key Concepts

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

Convergence
Convergence is a fundamental idea in the study of series. In a mathematical context, a series converges if the sequence of its partial sums approaches a finite limit. Consider a series \( \sum_{n=0}^{\infty} a_n \), which is said to converge to the sum \( S \) if, as \( n \) increases, the sum of the first \( n \) terms approaches \( S \). This happens when the differences between successive partial sums become negligible in the limit. In the given problem, the series \( \sum_{n=0}^{\infty} \frac{2^n - 1}{3^n} \) is split into two separate geometric series. By analyzing its convergence through the ratio \( r \) of the series, which must satisfy \( |r| < 1 \), both parts of the series were determined to converge. The fact that each part converges individually ensures that the original series also converges.
Summation
Summation, represented by the symbol \( \sum \), is the process of adding a sequence of numbers. The process of summation is vital when dealing with series, especially when we want to find the total value of the series. In the context of geometric series, this process becomes straightforward with a known formula.For a geometric series with the first term \( a \) and common ratio \( r \), the sum of an infinite series is given by the formula: - \( \frac{a}{1 - r} \) provided \( |r| < 1 \). In the step-by-step solution, the original series was split into two geometric series. Each of these was summed separately using this key formula. The accuracy of summation lies in ensuring that the common ratio \( r \) is indeed less than 1, thereby assuring convergence and allowing us to find a finite sum.
Series Divergence
While convergence pertains to a series settling on a particular sum, divergence refers to the series failing to approach a limit. When analyzing series, it is crucial to ascertain whether a series is converging or diverging because it helps in understanding the behavior and characteristics of the series.In our case, \( \sum_{n=0}^{\infty} \frac{2^n - 1}{3^n} \), we initially suspect divergence since the terms \( 2^n \) and \( 1 \) differ significantly. However, when broken into smaller parts, each behaves as a geometric series with satisfactory ratios imposed: \( \frac{2}{3} \) and \( \frac{1}{3} \), both ensuring convergence, hence showing no divergence. By understanding divergence, one ensures not to make unfounded conclusions about the series’ tendencies.
Mathematical Analysis
Mathematical analysis involves a methodical approach to understanding the properties and behaviors of series, functions, and generally infinite processes. It combines various techniques and doctrines to evaluate functions and the sequence of their sums.In this exercise, mathematical analysis was applied when decomposing the series \( \sum_{n=0}^{\infty} \frac{2^n - 1}{3^n} \) into two geometric series. This analytical approach allowed a clear path to examine each portion separately and apply known convergent criteria. Such analysis confirms convergence properties, enables calculation of sum totals, and prevents errors like misjudging divergence. - By exploring the ratios and applying geometric forms, mathematical analysis helps not just in solving series problems but also in developing a deeper intuition about infinite series and function behavior.

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Most popular questions from this chapter

In Exercises \(33-36\) , use series to estimate the integrals' values with an error of magnitude less than \(10^{-3} .\) (The answer section gives the integrals' values rounded to five decimal places.) $$ \int_{0}^{0.1} \frac{1}{\sqrt{1+x^{4}}} d x $$

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+y=0, \quad y^{\prime}(0)=0 \text { and } y(0)=1 $$

Estimate the error if \(\cos t^{2}\) is approximated by \(1-\frac{t^{4}}{2}+\frac{t^{8}}{4 !}\) in the integral \(\int_{0}^{1} \cos t^{2} d t .\)

According to a front-page article in the December \(15,1992,\) issue of the Wall Street Journal, Ford Motor Company used about 7\(\frac{1}{4}\) hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in \(1980 .\) The Japanese needed only about 3\(\frac{1}{2}\) hours. Ford's improvement since 1980 represents an average decrease of 6\(\%\) per year. If that rate continues, then \(n\) years from 1992 Ford will use about $$ S_{n}=7.25(0.94)^{n} $$ hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3\(\frac{1}{2}\) hours per vehicle, how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence \(\left\\{S_{n}\right\\}\) that is less than or equal to \(3.5 .\) b. Graph \(f(x)=7.25(0.94)^{x}\) and use Trace to find where the graph crosses the line \(y=3.5 .\)

Euler's constant Graphs like those in Figure 11.8 suggest that as \(n\) increases there is little change in the difference between the sum $$1+\frac{1}{2}+\cdots+\frac{1}{n}$$ and the integral $$\ln n=\int_{1}^{n} \frac{1}{x} d x$$ To explore this idea, carry out the following steps. a. By taking \(f(x)=1 / x\) in the proof of Theorem 9 , show that $$\ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n$$ or $$0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1$$ Thus, the sequence $$ a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ is bounded from below and from above. b. Show that $$ \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n $$ and use this result to show that the sequence \(\left\\{a_{n}\right\\}\) in part (a) is decreasing. since a decreasing sequence that is bounded from below converges (Exercise 107 in Section 11.1\()\) , the numbers \(a_{n}\) defined in part (a) converge: $$1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma$$ The number \(\gamma,\) whose value is \(0.5772 \ldots,\) is called Euler's constant. In contrast to other special numbers like \(\pi\) and \(e,\) no other expression with a simple law of formulation has ever been found for \(\gamma .\)

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