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Chapter 8: Problem 32

Compute several partial sums of the series \(1+1-1+1-1+1-1+\cdots\) Argue that the limit of the sequence of partial sums does not exist, so that the series diverges. Also, write this series as a geometric series and use Theorem 2.1 to conclude that the series diverges. Finally, use the \(k\) th-term test for divergence to conclude that the series diverges.

Short Answer

Expert verified
The series \(1+1-1+1-1+1-1+\cdots\) diverges. This conclusion is based on the observation that the series' partial sums do not approach any limit, it does not satisfy the condition for an infinite geometric series to converge (Theorem 2.1), and it fails the \(k\)th-term test for divergence.

Step by step solution

01

Partial Sums

Compute the first few partial sums: \(S_1=1\), \(S_2=2\), \(S_3=1\), \(S_4=2\), etc. Observe that the partial sums do not approach any limit, but rather alternate between 1 and 2. Therefore, the limit of the sequence of partial sums does not exist.
02

Write as Geometric Series

The series can be rewritten as an infinite geometric series with first term (a) = 1 and common ratio (r) = -1. The resulting series is \(1+(-1)+1+(-1)+1+(-1)+\cdots\).
03

Apply Theorem 2.1

As per Theorem 2.1, an infinite geometric series converges if and only if the absolute value of the common ratio is less than 1. In this case, the common ratio is -1 and its absolute value is 1, which is not less than 1. Hence, as per Theorem 2.1, the series does not converge, i.e., the series diverges.
04

Apply the \(k\)th-Term Test for Divergence

The \(k\)th-term test for divergence states: if the limit as \(n\) approaches infinity of \(a_n\) does not equal 0, then the series \(\sum_{n=1}^{\infty} a_n\) diverges. In this series, the \(k\)th term does not approach 0, but alternates between 1 and -1. Therefore, by the \(k\)th-term test, the series diverges.

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Key Concepts

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

Partial Sums
When examining the convergence of a series, a fundamental concept is that of partial sums. Partial sums are simply the sum of the first n terms of a series. They are denoted as Sn, where n represents the number of terms added together.

For the series in question, the partial sums alternate as S1=1, S2=2, S3=1, and so on. The nature of a convergent series is such that as n grows larger, the partial sums should approach a fixed number, called the limit. However, in this case, the partial sums oscillate indefinitely between two values without settling down, meaning there is no such limit. Thus, by the behavior of its partial sums, the series is identified as divergent.
Geometric Series
A geometric series is one where each term after the first is found by multiplying the previous term by a constant called the common ratio (r). The form of a geometric series is a + ar + ar2 + ar3 + ..., where a is the first term.

The series 1+1-1+1-1+1-1+... can be written as 1 + (-1) + 1 + (-1) + 1 +..., with a = 1 and r = -1. This places it in the geometric series category, although it's an atypical one since the common ratio is negative, leading to the alternation in sign for subsequent terms.
Theorem 2.1
Theorem 2.1 is a criterion for the convergence of an infinite geometric series. It states that such a series converges if and only if the absolute value of the common ratio is less than 1. Applied to the current series, wherein the common ratio r is -1, we see that the absolute value of r is actually |-1| = 1, which does not satisfy the condition for convergence (|r| < 1).

Therefore, according to Theorem 2.1, the geometric series 1 - 1 + 1 - 1 + ... does not converge and is, in fact, divergent.
kth-Term Test for Divergence
The kth-Term Test is another valuable tool in analyzing series for convergence or divergence. It states that if the limit of the kth term of the series, as k approaches infinity, is not zero, then the series diverges.

In this case, the terms of the series are either 1 or -1, which does not approach zero as we move through successive terms. In essence, there's no approaching a single value; instead, the terms alternate without approaching any particular point. Thus, the series fails the kth-Term Test for Divergence and we can conclude decisively that it is divergent.

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Most popular questions from this chapter

We have seen that \(\sin 1=1-\frac{1}{31}+\frac{1}{51}+\cdots .\) Determine how many terms are needed to approximate sin 1 to within \(10^{-5}\) Show that \(\sin 1=\int_{0}^{1} \cos x d x .\) Determine how many points are needed for Simpson's Rule to approximate this integral to within \(10^{-5} .\) Compare the efficiency of Maclaurin series and Simpson's Rule for this problem.

This problem is sometimes called the coupon collectors" problem. The problem is faced by collectors of trading cards. If there are \(n\) different cards that make a complete set and you randomly obtain one at a time, how many cards would you expect to obtain before having a complete set? (By random, we mean that each different card has the same probability of \(\frac{1}{n}\) of being the next card obtained.) In exercises \(69-72,\) we find the answer for \(n=10 .\) The first step is simple; to collect one card you need to obtain one card. Now, given that you have one card, how many cards do you need to obtain to get a second (different) card? If you're lucky, the next card is it (this has probability 10). But your next card might be a duplicate, then you get a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{9}{10}\right) .\) Or you might get two duplicates and then a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}\right) ;\) and so on. The mean is \(1 \cdot \frac{9}{10}+2 \cdot \frac{1}{10} \cdot \frac{9}{10}+3 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}+\cdots\) or \(\sum_{k=1}^{\infty} k\left(\frac{1}{10}\right)^{k-1}\left(\frac{9}{10}\right)=\sum_{k=1}^{\infty} \frac{9 k}{10^{k}} .\) Using the same trick as in exercise \(68,\) show that this is a convergent series with \(\operatorname{sum} \frac{10}{9}\)

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Use a Taylor series to verify the given formula. $$\sum_{i=0}^{\infty} \frac{(-1)^{k} \pi^{2 k+1}}{(2 k+1) !}=0$$

Find the Taylor series of \(f(x)=x \ln x\) about \(c=1 .\) Compare to the Taylor series for \(\ln x\) about \(c=1\)
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