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Chapter 8: Problem 16

Show that the series diverges. \(1-\frac{3}{2}+\frac{9}{4}-\frac{27}{8}+\cdots\)

Short Answer

Expert verified
The given series is an alternating series that can be rewritten as \(\sum_{n=0}^\infty (-1)^n \frac{3^n}{2^n}\). However, the sequence \(a_n=\frac{3^n}{2^n}\) is not decreasing and doesn't converge to 0, as \(\lim_{n\to\infty} \left(\frac{3}{2}\right)^n\) doesn't converge to 0. Therefore, the series diverges.

Step by step solution

01

Understand the Series

The given series is: \(1-\frac{3}{2}+\frac{9}{4}-\frac{27}{8}+\cdots\) This is an alternating series, as the terms switch between positive and negative. To apply the Alternating Series Test, we need to rewrite the series in the following form: \(\sum_{n=1}^\infty (-1)^n a_n\) where \(a_n\) is a non-negative sequence that is decreasing and converging to 0.
02

Rewrite the Series

The given series can be rewritten as: \(\sum_{n=0}^\infty (-1)^n \frac{3^n}{2^n}\) It's easy to see that \(a_n=\frac{3^n}{2^n}\) and it is indeed a non-negative sequence.
03

Check If the Sequence a_n is Decreasing

To check whether the sequence \(a_n\) is decreasing, we can compare consecutive terms: \(\frac{a_{n+1}}{a_n} = \frac{3^{n+1} / 2^{n+1}}{3^n / 2^n} = \frac{3}{2}\) Since the ratio between consecutive terms is greater than 1, the sequence \(a_n\) is not decreasing.
04

Check If the Sequence a_n Converges to 0

To check for convergence, we look at the limit: \(\lim_{n\to\infty} \frac{3^n}{2^n}\) To solve this limit, we can rewrite it as: \(\lim_{n\to\infty} \left(\frac{3}{2}\right)^n\) This limit doesn't converge to 0, as the ratio \(\frac{3}{2}\) is greater than 1.
05

Conclusion

Since the sequence \(a_n\) is neither decreasing nor converging to 0, we conclude that the series diverges.

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Key Concepts

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

Divergence
In the realm of mathematical series, understanding whether a series diverges or converges is crucial. A series diverges if the sequence of its partial sums does not settle at a particular value as more terms are added. For the series in question \( 1 - \frac{3}{2} + \frac{9}{4} - \frac{27}{8} + \cdots \), we need to determine its behavior. In this case, the series diverges because the sequence \( a_n = \frac{3^n}{2^n} \) fails to meet necessary conditions for convergence. Specifically, it isn't decreasing, nor does it approach zero.
  • Divergence indicates that the series sums increase infinitely without approaching a finite number.
  • This can be due to the terms not diminishing or not settling into a pattern that converges.
  • The partial sums grow without bound and fail to stabilize around a particular value.
Decreasing Sequence
A decreasing sequence is one where each term is less than or equal to the preceding term. For a sequence \( a_n = \frac{3^n}{2^n} \), we examine the ratio of successive terms to determine if it is decreasing.
Checking this, we find \( \frac{a_{n+1}}{a_n} = \frac{3}{2} \), indicating that each term increases rather than decreases. A crucial requirement for the Alternating Series Test is having a decreasing sequence. Since our sequence does not decrease, we already stumble on a significant issue.
  • Decreasing sequences allow terms to lessen in size, contributing to eventual convergence.
  • If a sequence grows or remains constant, it poses a significant hurdle for convergence.
  • This is why we initially check whether our series is composed of a decreasing sequence.
Series Convergence
Convergence of a series refers to whether its partial sums approach a specific value as we add more terms. For the Alternating Series Test to conclude convergence, the series must meet a set of criteria. These include having a sequence that is decreasing and approaches zero. In our case, because the series \( \sum_{n=0}^\infty (-1)^n \frac{3^n}{2^n} \) fails to meet these requirements, it cannot converge.
  • Convergence indicates that a series' sums get closer to a definitive limit despite adding more terms.
  • In alternating series, convergence particularly requires a non-increasing sequence with terms trending toward zero.
  • Failure for a series to meet the convergence criteria means sums oscillate or grow infinitely, indicating divergence.
Non-Negative Sequence
A non-negative sequence consists entirely of non-negative numbers鈥攅ither positive or zero. This characteristic is imperative when applying the Alternating Series Test. In the example series characterized by \( a_n = \frac{3^n}{2^n} \), we see that \( a_n \) remains non-negative, as both the numerator and denominator are positive, ensuring non-negative outcomes.
  • Non-negative sequences ensure that terms added to a series do not "pull the sum down," typically a condition for tests involving series.
  • Keeping sequences non-negative ensures the proper application of convergence tests like the Alternating Series Test.
  • While the sequence \( a_n \) in our example is non-negative, it doesn't fulfill other criteria needed for proving convergence.

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Most popular questions from this chapter

Determine whether the series is convergent, absolutely convergent, conditionally convergent, or divergent. \(\sum_{n=2}^{\infty} \frac{\sin \left(\frac{n \pi}{4}\right)}{n(\ln n)^{2}}\)

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. If \(a_{n}>0\) for \(n \geq 1\) and \(\Sigma_{n=1}^{\infty} a_{n}\) converges, then \(\Sigma_{n=1}^{\infty}(-1)^{n} a_{n}\) converges.

a. Show that if \(\sum a_{n}\) converges absolutely, then \(\sum a_{n}^{2}\) converges. b. Show that the converse of the result in part (a) is false by finding a series \(\sum a_{n}\) for which \(\sum a_{n}^{2}\) converges, but \(\Sigma\left|a_{n}\right|\) diverges.

Use the power series representations of functions established in this section to find the Taylor series of \(f\) at the given value of \(c .\) Then find the radius of convergence of the series. \(f(x)=x^{2} \cos x, \quad c=0\)

Evaluate \(\lim _{x \rightarrow 1} \frac{\ln x}{x-1}\). Hint: Use the Taylor series representation of \(\ln x\) at \(1 .\)
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