/*! 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 3 Determine convergence or diverge... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 3

Determine convergence or divergence of the series. $$\sum_{k=4}^{\infty} k^{-11 / 10}$$

Short Answer

Expert verified
The series \(\sum_{k=4}^{\infty} k^{-11 / 10}\) is convergent.

Step by step solution

01

Identify series type

The given series is \(\sum_{k=4}^{\infty} k^{-11 / 10}\). It is a slight variation of a p-series. Where instead of running the series from k = 1, it runs from k = 4 but the same principle will still apply.
02

Apply the p-series test

The p-series test for convergence states that if we have a series in the form \(\sum_{k=1}^{\infty} k^{-p}\), it will converge if \(p > 1\). In case of this series, the p value is \(11 / 10\) which is greater than 1.
03

Determine convergence or divergence

Since the p value is greater than 1, according to the p-series test, the series \(\sum_{k=4}^{\infty} k^{-11 / 10}\) will converge.

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!

Key Concepts

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

Understanding the p-series test
The p-series test is a fundamental concept in calculus series, particularly helpful in determining whether a series converges or diverges. A p-series is a series of the form \[ \sum_{k=1}^{\infty} \frac{1}{k^p} \]where \( p \) is a positive constant. To determine the convergence of a p-series, you need to check the value of \( p \).

Here is how the test works:
  • If \( p > 1 \), the series converges. This is because the terms decrease and sum up to a finite number.
  • If \( p \leq 1 \), the series diverges. As the terms do not decrease fast enough, they grow unbounded.
In our exercise, we dealt with the series given as \[ \sum_{k=4}^{\infty} k^{-11 / 10} \]Here, the term \( k^{-11 / 10} \) can be re-written as \( \frac{1}{k^{11/10}} \), which fits the structure of a p-series with \( p = \frac{11}{10} \). Since \( 11/10 > 1 \), the series converges.
Basics of Series Divergence
Series divergence refers to a situation where the series fails to add up to a finite number as more terms are considered. Unlike convergent series, diverging series continue indefinitely without settling into a finite sum.

In the context of the exercise, divergence occurs if the conditions of the p-series test are not met, specifically if \( p \leq 1 \).
  • If \( p = 1 \), the series becomes the harmonic series, notorious for its divergence, as its terms decrease too slowly.
  • If \( p < 1 \), the divergence is even more pronounced, because the terms decrease even more slowly or possibly increase.

Knowing whether a series converges or diverges is crucial in calculus since it determines whether resulting sums are finite and meaningful. In our exercise, confirming that \( 11/10 > 1 \) reassures us that divergence does not occur.
Introduction to Calculus Series
In calculus, series are infinite sums of terms arranged in a sequence. Each term is typically generated by applying a function to successive integer inputs. Understanding series is pivotal for solving numerous real-world problems through calculus.

There are different types of series you might encounter, each with unique behaviors and tests for convergence or divergence:
  • Geometric series: A series where each term is obtained by multiplying the previous term by a constant. For example, \( a + ar + ar^2 + \ldots \).
  • Harmonic series: A divergent series of the form \( \sum \frac{1}{n} \).
  • p-series: Series of the form \( \sum_{k=1}^{\infty} \frac{1}{k^p} \). Tested using the p-series test discussed earlier.

In studying these series, one often examines aspects such as convergence, divergence, and the rate at which terms shrink. Mastering these concepts allows students to tackle complex problems involving sums that stretch to infinity, a cornerstone of calculus.

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

Suppose that an astronaut is at (0,0) and the moon is represented by a circle of radius 1 centered at \((10,5) .\) The astronaut's capsule follows a path \(y=f(x)\) with current position \(f(0)=0,\) slope \(f^{\prime}(0)=1 / 5,\) concavity \(f^{\prime \prime}(0)=-1 / 10\) \(f^{\prime \prime \prime}(0)=1 / 25, f^{(4)}(0)=1 / 25\) and \(f^{(5)}(0)=-1 / 50 .\) Graph a Taylor polynomial approximation of \(f(x) .\) Based on your current information, do you advise the astronaut to change paths? How confident are you in the accuracy of your approximation?

Determine the radius and interval of convergence. $$\sum_{k=4}^{\infty} \frac{1}{k^{2}}(x+2)^{k}$$

Find all values of \(p\) such that the sequence \(a_{n}=\frac{1}{n^{p}}\) converges.

Determine the radius and interval of convergence. $$\sum_{k=1}^{\infty} \frac{4^{k}}{\sqrt{k}} x^{k}$$

Determine the radius and interval of convergence. $$\sum_{i=0}^{\infty} \frac{3^{k}}{k !} x^{k}$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

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.