/*! 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 93 State the Ratio Test.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

State the Ratio Test.

Short Answer

Expert verified
The Ratio Test states that for a series \(\sum_{n=1}^{\infty} a_n\), let \(L = \lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}\). If \(L < 1\), the series is absolutely convergent. If \(L > 1\), the series is divergent. If \(L = 1\) or the limit does not exist, the test is inconclusive.

Step by step solution

01

Define the Ratio Test

First, introduce the Ratio Test. The Ratio Test is a method used in mathematics, particularly in calculus, to determine whether a series converges or diverges.
02

State and Explain the Ratio Test

The Ratio Test states that for a series \(\sum_{n=1}^{\infty} a_n\), where \(a_n\) are positive terms, let \(L = \lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}\). If \(L < 1\), then the series \(\sum_{n=1}^{\infty} a_n\) is absolutely convergent. If \(L > 1\), then the series \(\sum_{n=1}^{\infty} a_n\) is divergent. If \(L = 1\) or the limit does not exist, then the test is inconclusive - the series may either be divergent or convergent.
03

Implication of the Ratio Test

Every series will fall into one of the three categories on the basis of this test. A series is said to be convergent if it approaches a particular number and divergent if it does not. If the limit equals 1 or does not exist, then the test does not provide any information.

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

Projectile Motion A projectile fired from the ground follows the trajectory given by $$y=\left(\tan \theta-\frac{g}{k v_{0} \cos \theta}\right) x-\frac{g}{k^{2}} \ln \left(1-\frac{k x}{v_{0} \cos \theta}\right)$$ where \(v_{0}\) is the initial speed, \(\theta\) is the angle of projection, \(g\) is the acceleration due to gravity, and \(k\) is the drag factor caused by air resistance. Using the power series representation $$\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots, \quad-1 < x < 1$$ verify that the trajectory can be rewritten as $$y=(\tan \theta) x+\frac{g x^{2}}{2 v_{0}^{2} \cos ^{2} \theta}+\frac{k g x^{3}}{3 v_{0}^{3} \cos ^{3} \theta}+\frac{k^{2} g x^{4}}{4 v_{0}^{4} \cos ^{4} \theta}+\cdots$$

Finding a Taylor Series In Exercises \(1-12,\) use the definition of Taylor series to find the Taylor series, centered at \(c,\) for the function. $$ f(x)=\sin x, \quad c=\frac{\pi}{4} $$

Approximating an Integral In Exercises \(63-70\) , use a power series to approximate the value of the integral with an error of less than \(0.0001 .\) (In Exercises 65 and \(67,\) assume that the integrand is defined as 1 when \(x=0 .\) $$ \int_{0}^{0.2} \sqrt{1+x^{2}} d x $$

Finding a Maclaurin Series In Exercises \(41-44\) , find the Maclaurin series for the function. (See Examples 7 and \(8 . )\) $$ f(x)=x \sin x $$

Finding a Limit In Exercises \(59-62,\) use the series representation of the function \(f\) to find \(\lim _{x \rightarrow 0} f(x)\) (if it exists). $$ f(x)=\frac{1-\cos x}{x} $$

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.