/*! 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 15 Use the Ratio Test to determine ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the Ratio Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{2^{k}}{k^{99}}$$

Short Answer

Expert verified
Question: Determine if the series $$\sum_{k=1}^{\infty} \frac{2^{k}}{k^{99}}$$ converges using the Ratio Test. Answer: The series $$\sum_{k=1}^{\infty} \frac{2^{k}}{k^{99}}$$ diverges.

Step by step solution

01

Identify the terms of the series

The terms of the series can be represented by the general term $$a_k = \frac{2^k}{k^{99}}$$
02

Apply the Ratio Test by finding the limit

Using the Ratio Test, we need to find the limit as k approaches infinity of the absolute ratio of consecutive terms, which is given by: $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| = \lim_{k\to\infty} \frac{\frac{2^{k+1}}{(k+1)^{99}}}{\frac{2^k}{k^{99}}}$$
03

Simplify the expression

Simplify the expression by eliminating common factors: $$\lim_{k\to\infty} \frac{2^{k+1}}{(k+1)^{99}}\cdot\frac{k^{99}}{2^k} = \lim_{k\to\infty} \frac{2}{\left(\frac{k+1}{k}\right)^{99}}$$
04

Use limit properties to evaluate the expression

Using the limit properties, we get $$\lim_{k\to\infty} \frac{2}{\left(\frac{k+1}{k}\right)^{99}} = \frac{2}{\left(1+\frac{1}{k}\right)^{99}}$$ Since the denominator goes to \(1^{\infty}\) as k approaches infinity, the limit of the absolute ratio of consecutive terms is: $$\lim_{k\to\infty} \frac{2}{\left(1+\frac{1}{k}\right)^{99}}=2$$
05

Compare the limit with 1

Since the calculated limit is greater than 1, $$\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|=2 > 1$$
06

Conclude about the convergence

With the Ratio Test, when the limit is greater than 1, we can conclude that the series diverges. Therefore, the series $$\sum_{k=1}^{\infty} \frac{2^{k}}{k^{99}}$$ diverges.

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!

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

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. The Consumer Price Index (the CPI is a measure of the U.S. cost of living) is given a base value of 100 in the year \(1984 .\) Assume the CPI has increased by an average of \(3 \%\) per year since \(1984 .\) Let \(c_{n}\) be the CPI \(n\) years after \(1984,\) where \(c_{0}=100.\)

Determine whether the following statements are true and give an explanation or counterexample. a. A series that converges must converge absolutely. b. A series that converges absolutely must converge. c. A series that converges conditionally must converge. d. If \(\sum a_{k}\) diverges, then \(\Sigma\left|a_{k}\right|\) diverges. e. If \(\sum a_{k}^{2}\) converges, then \(\sum a_{k}\) converges. f. If \(a_{k}>0\) and \(\sum a_{k}\) converges, then \(\Sigma a_{k}^{2}\) converges. g. If \(\Sigma a_{k}\) converges conditionally, then \(\Sigma\left|a_{k}\right|\) diverges.

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(R_{n}<10^{-6}\) ). a. \(\sum_{k=0}^{\infty} 0.72^{k}\) b. \(\sum_{k=0}^{\infty}(-0.25)^{k}\)

After many nights of observation, you notice that if you oversleep one night you tend to undersleep the following night, and vice versa. This pattern of compensation is described by the relationship $$x_{n+1}=\frac{1}{2}\left(x_{n}+x_{n-1}\right), \quad \text { for } n=1,2,3, \ldots$$ where \(x_{n}\) is the number of hours of sleep you get on the \(n\) th night and \(x_{0}=7\) and \(x_{1}=6\) are the number of hours of sleep on the first two nights, respectively. a. Write out the first six terms of the sequence \(\left\\{x_{n}\right\\}\) and confirm that the terms alternately increase and decrease. b. Show that the explicit formula $$x_{n}=\frac{19}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)^{n}, \text { for } n \geq 0,$$ generates the terms of the sequence in part (a). c. What is the limit of the sequence?

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}}=\frac{7 \pi^{4}}{720}.\) (Assume the result of Exercise 63.)

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.