/*! 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 26 Use the Root Test to determine w... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 26

Use the Root Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{k}{e^{k}}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Understand the Root Test

The Root Test can be used to determine if a series converges or diverges. The test involves finding the following limit: $$L = \lim_{n\to\infty} \sqrt[n]{|a_n|}$$ where \(a_n\) are the terms of the series. If \(L < 1\), the series converges; if \(L > 1\), the series diverges; if \(L = 1\), the test is inconclusive. In our case, we have: $$a_k = \frac{k}{e^k}$$ We need to find the limit of the kth root of the absolute value of this term as k goes to infinity.
02

Calculate the Limit of the kth Root

Now, we calculate the limit of the kth root of the absolute value of the term as k goes to infinity: $$L = \lim_{k\to\infty} \sqrt[k]{\left|\frac{k}{e^k}\right|} = \lim_{k\to\infty} \sqrt[k]{\frac{k}{e^k}}$$ To solve this limit, we can make use of the properties of exponentials and rewrite it as: $$L = \lim_{k\to\infty} \frac{\sqrt[k]{k}}{\sqrt[k]{e^k}}$$ $$L = \lim_{k\to\infty} \frac{\sqrt[k]{k}}{e}$$
03

Determine Convergence Using the Root Test Criteria

Using the Root Test criteria, the series converges if the limit is less than 1, diverges if the limit is greater than 1, or the test is inconclusive if the limit is equal to 1. We know that the limit: $$L = \lim_{k\to\infty} \frac{\sqrt[k]{k}}{e}$$ As k approaches infinity, the kth root of k (the numerator) approaches 1, while the denominator remains constant as e. Thus, the limit is: $$L = \frac{1}{e}$$ Since \(0 < \frac{1}{e} < 1\), the Root Test tells us that the given series converges.

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 91影视!

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

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{p^{k} k !}$$

Evaluate the limit of the following sequences. $$a_{n}=\frac{6^{n}+3^{n}}{6^{n}+n^{100}}$$

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5, n=0,1,2, \dots$$

Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty}(-1)^{k}$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation
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.