/*! 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 20 Determine whether the following ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine whether the following series converge. $$\sum_{k=0}^{\infty}\left(-\frac{1}{5}\right)^{k}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Identify the type of series

The series can be expressed as follows: $$\sum_{k=0}^{\infty}\left(-\frac{1}{5}\right)^{k}$$ This series is a geometric series because it has a constant ratio between consecutive terms.
02

Find the common ratio

The common ratio for a geometric series can be found by looking at the term that is being raised to the power of \(k\). In our case, the common ratio is: $$r = -\frac{1}{5}$$
03

Determine convergence using the geometric series test

For a geometric series with ratio \(r\), the series converges if \(|r| < 1\). Otherwise, the series diverges. In our case, we have the following: $$|r| = \left|-\frac{1}{5}\right| = \frac{1}{5}$$ Since \(\frac{1}{5} < 1\), the series converges.
04

Conclusion

Using the geometric series test, we determined that the series: $$\sum_{k=0}^{\infty}\left(-\frac{1}{5}\right)^{k}$$ converges because the common ratio is \(r = -\frac{1}{5}\) and \(|r| = \frac{1}{5} < 1\).

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

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

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

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

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

A well-known method for approximating \(\sqrt{c}\) for positive real numbers \(c\) consists of the following recurrence relation (based on Newton's method). Let \(a_{0}=c\) and $$a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right), \quad \text { for } n=0,1,2,3, \dots$$ a. Use this recurrence relation to approximate \(\sqrt{10} .\) How many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.01 ?\) How many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.0001 ?\) (To compute the error, assume a calculator gives the exact value.) b. Use this recurrence relation to approximate \(\sqrt{c},\) for \(c=2\) \(3, \ldots, 10 .\) Make a table showing how many terms of the sequence are needed to approximate \(\sqrt{c}\) with an error less than \(0.01.\)

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.