Answered: 1 Special series Does the series Σ Vk²… | bartleby

Calculus: Early Transcendentals

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

1
Special series Does the series
Σ
Vk² – 6k + 9
converge or diverge?
k=4
-

Expert Solution

Step 1

Given

$\sum_{k = 4}^{\infty} \frac{1}{\sqrt[4]{k^{2} - 6 k + 9}}$

Step 2

To determine whether the series is converge or diverge.

The integral test is,

Suppose f(x) is a positive decreasing continuous function on the interval $[1, \infty)$ with f(n)=a_n.

$T h e n t h e s e r i e s, \sum_{n = 1}^{\infty} a_{n} i s c o n v e r g e n t i f a n d o n l y i f \int_{1}^{\infty} f (x) d x c o n v e r g e s . T h a t i s i f \int_{1}^{\infty} f (x) d x i s c o n v e r g e n t, t h e n \sum_{n = 1}^{\infty} a_{n} i s c o n v e r g e n t . i f \int_{1}^{\infty} f (x) d x i s d i v e r g e n t, t h e n \sum_{n = 1}^{\infty} a_{n} i s d i v e r g e n t .$

Step 3

Here $a_{k} = \sqrt[4]{k^{2} - 6 k + 9}$

To determine whether the series converges or diverges we need to find $\int_{1}^{\infty} f (x) d x$ .

Here f(x)= $\frac{1}{\sqrt[4]{k^{2} - 6 k + 9}}$ .

To find these integral use p test for convergence which is,

$\int_{1}^{\infty} \frac{1}{x^{p}} d x c o n v e r g e s i f p > 1 a n d d i v e r g e s i f p \leq 1 .$

Consider

$\int_{1}^{\infty} \frac{1}{x^{p}} d x = \int_{1}^{\infty} \frac{1}{\sqrt[4]{k^{2} - 6 k + 9}} = \int_{1}^{\infty} \frac{1}{(k}$ ²-6k+9)1/4

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