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Chapter 9: Problem 15

Determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{3}}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Identify the Alternating Series

The given series is: $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{3}}$$ We can see that this is an alternating series, because the terms switch between positive and negative values due to the \((-1)^{k+1}\) part.
02

Split into two parts

To perform the Alternating Series Test, split the series into two parts: the alternating part and the non-alternating part. In this case, the alternating part is \((-1)^{k+1}\), and the non-alternating part is \(\frac{1}{k^{3}}\).
03

Check if the non-alternating part is decreasing

The first condition of the Alternating Series Test says that the non-alternating part (\(\frac{1}{k^{3}}\)) must be a decreasing sequence. In this case, we can show that taking the derivative of the non-alternating part with respect to k will result in a negative value: $$\frac{d}{dk}\left(\frac{1}{k^{3}}\right) = -\frac{3}{k^{4}}$$ Since the derivative is always negative for \(k\geq1\), the non-alternating sequence is decreasing.
04

Check if the non-alternating part approaches zero

The second condition of the Alternating Series Test requires that the non-alternating part approaches zero as \(k\) approaches infinity: $$\lim_{k\to\infty}\frac{1}{k^{3}}=0$$ This limit does indeed equal zero.
05

Apply the Alternating Series Test

Both conditions of the Alternating Series Test are met: 1. The non-alternating part is a decreasing sequence. 2. The non-alternating part approaches zero as \(k\) approaches infinity. Therefore, we can conclude that the given alternating series converges.

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Most popular questions from this chapter

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{2^{k}}$$

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}}\). When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots $$ Use estimation techniques to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

a. Sketch the function \(f(x)=1 / x\) on the interval \([1, n+1]\) where \(n\) is a positive integer. Use this graph to verify that $$ \ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<1+\ln n $$ b. Let \(S_{n}\) be the sum of the first \(n\) terms of the harmonic series, so part (a) says \(\ln (n+1)0,\) for \(n=1,2,3, \ldots\) c. Using a figure similar to that used in part (a), show that $$ \frac{1}{n+1}>\ln (n+2)-\ln (n+1) $$ d. Use parts (a) and (c) to show that \(\left\\{E_{n}\right\\}\) is an increasing sequence \(\left(E_{n+1}>E_{n}\right)\) e. Use part (a) to show that \(\left\\{E_{n}\right\\}\) is bounded above by 1 f. Conclude from parts (d) and (e) that \(\left\\{E_{n}\right\\}\) has a limit less than or equal to \(1 .\) This limit is known as Euler's constant and is denoted \(\gamma\) (the Greek lowercase gamma). g. By computing terms of \(\left\\{E_{n}\right\\},\) estimate the value of \(\gamma\) and compare it to the value \(\gamma \approx 0.5772 .\) (It has been conjectured, but not proved, that \(\gamma\) is irrational.) h. The preceding arguments show that the sum of the first \(n\) terms of the harmonic series satisfy \(S_{n} \approx 0.5772+\ln (n+1)\) How many terms must be summed for the sum to exceed \(10 ?\)

Evaluate the limit of the following sequences. $$a_{n}=\frac{4^{n}+5 n !}{n !+2^{n}}$$

Series of squares Prove that if \(\sum a_{k}\) is a convergent series of positive terms, then the series \(\Sigma a_{k}^{2}\) also converges.
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