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In Exercises \(1-6,\) find a formula for the \(n\) th partial sum of each series and use it to find the series' sum if the series converges. $$ 1-2+4-8+\cdots+(-1)^{n-1} 2^{n-1}+\cdots $$

Short Answer

Expert verified
The series does not converge because the common ratio is greater than 1.

Step by step solution

01

Identify the Series Pattern

The given series is \(1 - 2 + 4 - 8 + \cdots + (-1)^{n-1} 2^{n-1} + \cdots\). Each term can be written in the form \((-1)^{n-1} 2^{n-1}\), where the power of 2 increases by 1 for each term, and the sign alternates between positive and negative.
02

Write the General Formula for the nth Term

The general formula for the \(n\)th term of the series is \(a_n = (-1)^{n-1} 2^{n-1}\). This accounts for both the changing signs and the exponentially increasing value.
03

Find the Formula for the nth Partial Sum

The \(n\)th partial sum \(S_n\) of the series is given by \(S_n = \sum_{k=1}^{n} (-1)^{k-1} 2^{k-1}\). To find this, we recognize the series as a modified geometric series with first term 1 and common ratio -2.
04

Use the Geometric Series Formula

For a geometric series, the sum of the first \(n\) terms is \(S_n = a\frac{1-r^n}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio. Here, \(a = 1\) and \(r = -2\), so: \[ S_n = 1 \frac{1 - (-2)^n}{1 - (-2)} = \frac{1 - (-2)^n}{3}. \]
05

Check for Convergence

A series converges if the absolute value of the common ratio \(|r| < 1\). Here, \(|r| = 2\), which is greater than 1. Therefore, the series does not converge.

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Key Concepts

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

Partial Sum
In a geometric series, understanding the concept of a partial sum is crucial. A partial sum, denoted as \(S_n\), is the sum of the first \(n\) terms of the series. For the given series, the general formula for the partial sum is derived from the pattern we observe among the terms. Clearly, the series is alternating, and each term is of the form \((-1)^{n-1} 2^{n-1}\).
To find the partial sum, we identify this series as a geometric sequence with the sequence's terms multiplied by a specific factor, in this case, \(-2\). Using the geometric series formula:
  • The first term \(a = 1\).
  • The common ratio \(r = -2\).
Applying these into the geometric formula for the partial sum, \(S_n = a \frac{1 - r^n}{1 - r}\), helps process the summation from the first term up to the \(n\)th term. Hence, \(S_n = \frac{1 - (-2)^n}{3}\).
This formula simplifies the process of calculating the sum of terms up to any given point \(n\), which is particularly useful when dealing with extensive series.
Convergent Series
A convergent series is one where the infinite sum approaches a finite limit as more terms are added. The convergence of a series is heavily dependent on the absolute value of the common ratio \(|r|\). If \(|r| < 1\), the series converges; conversely, if \(|r| \geq 1\), the series diverges.
When considering our series, the common ratio is \(-2\). The absolute value \(|-2| = 2\), which is greater than 1, indicating that the series diverges rather than converges. In simple terms, as you keep adding more terms of this particular series, instead of getting closer to a specific number and stabilizing, the sum will oscillate wildly without ever settling down.
It's important to recognize convergent series in mathematical analysis as they offer insights into stability and comprehensible limits, which are not present here due to the non-convergent nature of the series in question.
General Term Formula
To analyze any series, it's vital first to determine the general term, denoted as \(a_n\). This term helps in understanding the overall structure and behavior of the series. For our series, each term is formulated as \((-1)^{n-1} 2^{n-1}\).
This formula indicates two things:
  • The coefficients alternate in sign based on the term index \(n\), provided by \((-1)^{n-1}\).
  • The exponential growth factor \(2^{n-1}\), shows how each term increases in size.
Understanding the general term goes beyond identifying the individual components. It provides all the necessary details to reconstruct each specific term of the series and aids in pattern recognition, which is foundational when calculating partial sums and examining convergence traits.
By clearly identifying the general term, you set the groundwork to determine other valuable elements like the partial sum and whether or not the series converges or diverges.

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Most popular questions from this chapter

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a.\) find the (a) linearization (Taylor polynomial of order 1) and (b) quadratic approximation of \(f\) at \(x=0\). \(f(x)=\ln (\cos x)\)

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.

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} $$f(x)=e^{-x} \cos 2 x, \quad|x| \leq 1$$

Show that the Taylor series for \(f(x)=\tan ^{-1} x\) diverges for \(|x|>1 .\)

When \(a\) and \(b\) are real, we define \(e^{(a+i b) x}\) with the equation \begin{equation} e^{(a+i b) x}=e^{a x} \cdot e^{i b x}=e^{a x}(\cos b x+i \sin b x) \end{equation} Differentiate the right-hand side of this equation to show that \begin{equation} \frac{d}{d x} e^{(a+i b) x}=(a+i b) e^{(a+i b) x} \end{equation} Thus the familiar rule \((d / d x) e^{k x}=k e^{k x}\) holds for \(k\) complex as well as real.

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