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Chapter 11: Problem 4

Find a formula for the \(n\)th partial sum of each series and use it to find the series鈥� 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, so it does not have a sum.

Step by step solution

01

Identify the Pattern of the Series

The given series is: \(1 - 2 + 4 - 8 + \cdots\ (-1)^{n-1} 2^{n-1} + \cdots\). Each term in the series can be expressed as \((-1)^{n-1} \cdot 2^{n-1}\), where the exponent of -1 indicates the alternating sign and the exponent of 2 represents the successive power.
02

Write the General Term for the Series

From Step 1, the general term of the series is given by \(a_n = (-1)^{n-1} \cdot 2^{n-1}\), where \(n\) starts from 1 and goes to infinity.
03

Derive the Formula for the Partial Sum

The \(n\)th partial sum \(S_n\) of the series is the sum of the first \(n\) terms. Thus, the partial sum is given by \(S_n = \sum_{k=1}^{n} (-1)^{k-1} \cdot 2^{k-1} \).Recognize the geometric series component: \((-1)^{k-1} \cdot 2^{k-1}\) describes a geometric series with a common ratio \(r = -2\).
04

Calculate the Partial Sum for Geometric Series

For a geometric series with first term \(a = 1\) and common ratio \(r = -2\), the sum of the first \(n\) terms is given by the formula for geometric series:\[ S_n = a \frac{1-r^n}{1-r} \] Substituting \(a=1\) and \(r=-2\), we get:\[ S_n = \frac{1 - (-2)^n}{1 - (-2)} = \frac{1 - (-2)^n}{3} \].
05

Determine Convergence of the Series

To determine if the series converges, we consider the behavior of \((-2)^n\) as \(n\) approaches infinity. Since \(|r| = 2 > 1\), the series does not converge as \((-2)^n\) diverges to \(\pm \infty\).
06

Conclude with the Sum of the Series

Since the series does not converge due to the increasing magnitude of \((-2)^n\), it does not have a finite sum. Therefore, the sum of the series cannot be determined.

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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 sequence of numbers where each term after the first is found by multiplying the previous term by a constant, called the "common ratio." In this particular problem, the given series is recognized as a geometric series because of the pattern in its general term, \ \[a_n = (-1)^{n-1} \cdot 2^{n-1}\]. The common ratio \(r\) between consecutive terms is \(-2\). This means each term is \(-2\) times the previous term, resulting in an alternating sign and doubling the value with each step.
  • The *first term* \(a\) is \(1\).
  • The *common ratio* \(r\) is \(-2\).
Understanding the geometric series helps in identifying the formula for the partial sum, which is crucial for determining the sum of the series if it converges.
Series Convergence
Series convergence determines whether the sequence of partial sums of a series approaches a specific value as the number of terms goes to infinity. For geometric series, there is a special rule to test convergence: if the absolute value of the common ratio \(|r|<1\), the series converges.
However, in this exercise, we identified the common ratio as \(-2\). The absolute value \(|r|=2\), which is greater than 1, indicating that the series will not converge.
  • A series does not converge if its terms do not approach zero.
  • Instead, the terms grow larger or oscillate indefinitely, leading to divergence.
Thus, due to this divergence, there is no finite sum for this series.
General Term
The general term of a series gives you a formula to find any term in the series based on its position. For the series in this problem, the general term is \[a_n = (-1)^{n-1} \cdot 2^{n-1}\]. This expression reveals two things:
  • The *alternating sign* is given by \((-1)^{n-1}\), which flips every time \(n\) increments.
  • The *size of the term* grows exponentially based on \(2^{n-1}\).
Understanding the general term is crucial for recognizing patterns and computing sums for both partial and full series."
Using the general term, you can derive the formula for the partial sums, helping to determine whether the series converges or not.

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Most popular questions from this chapter

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+x^{2} y=x, \quad y^{\prime}(0)=b \text { and } y(0)=a $$

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{x \rightarrow \infty}(x+1) \sin \frac{1}{x+1} $$

Show by example that \(\sum_{n=1}^{\infty} a_{n} b_{n}\) may diverge even if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge.

Prove that \(\lim _{n \rightarrow \infty} \sqrt[n]{n}=1\).
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