/*! 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 52 Determine whether the following ... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the following series converge absolutely, converge conditionally, or diverge. $$\sum_{k=1}^{\infty}(-1)^{k} e^{-k}$$

Short Answer

Expert verified
Answer: The series converges absolutely.

Step by step solution

01

Check for Absolute Convergence

First, we will look into the absolute value of the series, thereby ignoring the (-1)^k factor: $$\sum_{k=1}^{\infty}|(-1)^{k} e^{-k}| = \sum_{k=1}^{\infty} e^{-k}$$ Now, we will determine whether this series converges or diverges. By doing so, we can either conclude whether the original series converges absolutely, or move onto the next step in case the result is inconclusive.
02

Apply the Ratio Test

In this step, we will apply the Ratio Test to the absolute series to ascertain its convergence. If the limit (L) of the ratio of consecutive terms in the series is less than 1, the series converges; if L is greater than 1, the series diverges. If L equals 1, the result is inconclusive. To proceed, we calculate the limit of the ratio: $$\lim_{k\to\infty}\frac{e^{-(k+1)}}{e^{-k}}$$
03

Simplify the Ratio

Simplify the ratio by dividing the exponents of e: $$\lim_{k\to\infty}\frac{e^{-(k+1)}}{e^{-k}} = \lim_{k\to\infty} e^{-1} = e^{-1}$$
04

Analyze the Result

Now that we have calculated e^{-1} (which is less than 1), the Ratio Test indicates that the absolute series converges, meaning the original series converges absolutely: $$\sum_{k=1}^{\infty}(-1)^{k} e^{-k}$$ This series converges absolutely.

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Key Concepts

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

Absolute Convergence
When studying infinite series, it's crucial to understand the concept of absolute convergence. An infinite series is said to be absolutely convergent if the series of absolute values of its terms converges.

For example, consider the series \(\sum_{k=1}^{\infty}(-1)^{k} e^{-k}\). To examine absolute convergence, we take the absolute value of each term, which leads us to the series \(\sum_{k=1}^{\infty} e^{-k}\). By investigating this new series, we can determine if the original series converges absolutely.

If the series of absolute values is convergent, then the original series will also converge. This holds regardless of the signs of the terms in the original series. In the case of our original series, if \(\sum_{k=1}^{\infty} e^{-k}\) converges, then \(\sum_{k=1}^{\infty}(-1)^{k} e^{-k}\) converges absolutely.
Ratio Test
The Ratio Test is a vital tool for determining the convergence of an infinite series. It involves examining the limit of the ratio of consecutive terms in the series.

To apply the Ratio Test, consider the limit \(\lim_{k\to\infty}\frac{a_{k+1}}{a_{k}}\), where \(a_{k}\) refers to the k-th term of the series. If this limit, commonly referred to as L, is less than 1, the series converges. If L is greater than 1, the series diverges, and if L equals 1, the test is inconclusive.

In the context of the series \(\sum_{k=1}^{\infty} e^{-k}\), we calculate the limit \(\lim_{k\to\infty}\frac{e^{-(k+1)}}{e^{-k}}\) which simplifies to \(e^{-1}\). Since \(e^{-1} < 1\), the Ratio Test tells us that the series converges, supporting the absolute convergence of our original series.
Conditional Convergence
Conditional convergence occurs when an infinite series converges, but it does not converge absolutely. That means that while the series \(\sum_{k=1}^{\infty}a_{k}\) might converge, the series of its absolute values \(\sum_{k=1}^{\infty}|a_{k}|\) diverges.

To illustrate, if we had discovered that our series \(\sum_{k=1}^{\infty}(-1)^{k} e^{-k}\) converged and the corresponding series of absolute values (the one without the alternating sign) diverged, we would then say that our series converges conditionally.

Conditional convergence is an interesting phenomenon because it explores the delicate nature of series where the alternating signs play a crucial role in the convergence of an otherwise divergent sequence of absolute terms.
Divergent Series
A divergent series is one that does not converge; that is, it does not approach a finite limit as more and more terms are added. The sum of the terms either increases indefinitely, decreases without bound, or oscillates without settling to a single value.

For instance, if in applying the Ratio Test to our series \(\sum_{k=1}^{\infty} e^{-k}\), we had found the limit L to be greater than 1, or if the series of absolute values \(\sum_{k=1}^{\infty}|(-1)^{k} e^{-k}|\) had diverged, we would conclude that our original series is divergent.

A divergent series does not possess a sum in the traditional sense, which distinguishes it markedly from a series that is absolutely or conditionally convergent. Understanding divergence is essential, as it indicates that the infinite sequence does not have a definite sum, often necessitating a closer look at the behavior of the individual terms.

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Most popular questions from this chapter

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{75^{n-1}}{99^{n}}+\frac{5^{n} \sin n}{8^{n}}$$

Consider the geometric series \(S=\sum_{k=0}^{\infty} r^{k}\) which has the value \(1 /(1-r)\) provided \(|r|<1\). Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The magnitude of the remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$ R_{n}=S-S_{n}=\frac{r^{n}}{1-r} $$

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The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

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