/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Determine whether the series is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine whether the series is absolutely convergent, conditionally convergent or divergent. $$\sum_{k=1}^{\infty} \frac{\cos k}{k^{3}}$$

Short Answer

Expert verified
The series \( \sum_{k=1}^{\infty} \frac{\cos k}{k^{3}} \) is absolutely convergent.

Step by step solution

01

Identify the series

The series is given by \( \sum_{k=1}^{\infty} \frac{\cos k}{k^{3}} \). To consider absolute convergence, the absolute value of the series terms must be taken.
02

Evaluate absolute convergence

The absolute sequence is \( \sum_{k=1}^{\infty} \left|\frac{\cos k}{k^{3}}\right| \). As \( |\cos k| \leq 1 \), this can be compared to \( \sum_{k=1}^{\infty} \frac{1}{k^{3}} \) which is a p-series with p>1, hence convergent. Thus, based on the comparison test, the original series is absolutely convergent.
03

Final conclusion

Since the series is absolutely convergent, we don't need to check for conditional convergence. Thus, the given series \( \sum_{k=1}^{\infty} \frac{\cos k}{k^{3}} \) is absolutely convergent.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Absolutely Convergent Series
An absolutely convergent series is one where the series of the absolute values of the terms converges. This means if we take all the negative signs away, and the series still converges, then it's absolutely convergent. This is a strong form of convergence. To determine if a series is absolutely convergent, we examine the absolute sum \[ \sum_{k=1}^{\infty} \left| a_k \right| \].

In our original series \( \sum_{k=1}^{\infty} \frac{\cos k}{k^{3}} \), we look at \( \sum_{k=1}^{\infty} \left|\frac{\cos k}{k^{3}}\right| \). Since \(|\cos k| \leq 1\), we can compare this to the series \( \sum_{k=1}^{\infty} \frac{1}{k^{3}} \). This series \( \frac{1}{k^{3}} \) is a p-series with \( p = 3 \), and since \( p > 1 \), the series converges. Hence, the original series is absolutely convergent.
Conditionally Convergent Series
While the given problem determines that the series is absolutely convergent, understanding what conditional convergence is can be helpful in many other contexts. A series is conditionally convergent if it converges only when you consider the original terms, not their absolute values. In simpler words, remove the negative signs, and it becomes divergent.

Mathematically, a series \( \sum_{k=1}^{\infty} a_k \) is conditionally convergent if the series itself converges, but the series of absolute values \( \sum_{k=1}^{\infty} \left| a_k \right| \) diverges.

For the series \( \sum_{k=1}^{\infty} \frac{\cos k}{k^{3}} \), the determination of its absolute convergence meant skipping this check, because absolutely convergent series are naturally convergent as well. This is something you always need to remember: absolute convergence implies regular convergence, but not vice versa.
Comparison Test
The comparison test is a handy tool for determining the convergence of a series. It involves comparing the series in question to another series that we already know converges or diverges.

To use this test, we find a series \( \sum_{k=1}^{\infty} b_k \) such that each term \( \left| a_k \right| \leq b_k \). If this comparison series converges and the terms are all non-negative, then the original series also converges. Similarly, if \( b_k \) diverges and \( a_k \geq b_k \) for large \( k \), our original series diverges.

In the problem, to evaluate absolute convergence, the absolute series \( \sum_{k=1}^{\infty} \left|\frac{\cos k}{k^{3}}\right| \) was compared to \( \sum_{k=1}^{\infty} \frac{1}{k^{3}} \), a known convergent p-series. Because \(|\cos k| \leq 1\), we know \( \left|\frac{\cos k}{k^{3}}\right| \leq \frac{1}{k^{3}} \), confirming the convergence of the original series.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the even/odd properties of \(f(x)\) to predict (don't compute) whether the Fourier series will contain only cosine terms, only sine terms or both. $$f(x)=|x|$$

If \(\sum_{k=0}^{\infty} a_{k} x^{k}\) has radius of convergence \(r,\) with \(0 < r < \infty,\) determine the radius of convergence of \(\sum_{k=0}^{\infty} a_{k}(x-c)^{k}\) for any constant \(c\)

Use the Binomial Theorem to approximate the value to within \(10^{-6}\) (a) \(\frac{2}{\sqrt[3]{9}}\) (b) \(\sqrt[4]{17}\)

Prove that if \(\sum_{k=0}^{\infty} a_{k} x^{k}\) has radius of convergence \(r,\) with \(0 < r < \infty,\) then \(\sum_{k=0}^{\infty} a_{k} x^{2 k}\) has radius of convergence \(\sqrt{r}\)

The function \(\sin 8 \pi t\) represents a \(4-\mathrm{Hz}\) signal \((1 \mathrm{Hz} \text { equals } 1\) cycle per second) if \(t\) is measured in seconds. If you received this signal, your task might be to take your measurements of the signal and try to reconstruct the function. For example, if you measured three samples per second, you would have the data \(f(0)=0, f(1 / 3)=\sqrt{3} / 2, f(2 / 3)=-\sqrt{3} / 2\) and \(f(1)=0\) Knowing the signal is of the form \(A\) sin \(B t,\) you would use the data to try to solve for \(A\) and \(B\). In this case, you don't have enough information to guarantee getting the right values for A and \(B\). Prove this by finding several values of \(A\) and \(B\) with \(B \neq 8 \pi\) that match the data. A famous result of \(\mathrm{H}\). Nyquist from 1928 states that to reconstruct a signal of frequency \(f\) you need at least \(2 f\) samples.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.