I need some help with this Calc problem 2. Consider that ) (x – 2)* is a geometric series. As such, for what values of x will this infinite seriesk-0converge? For those values of x, what will the series converge to (in terms of æ)?

Given that ∑k=0∞x-2kTo find : Value of xSolution :∑k=0∞x-2kRatio test ak=x-2k…

Answered: 2. Consider that > (x – 2)* is a…

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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I need some help with this Calc problem

2. Consider that ) (x – 2)* is a geometric series. As such, for what values of x will this infinite series
k-0
converge? For those values of x, what will the series converge to (in terms of æ)?

Expert Solution

Step 1

$G i v e n t h a t \sum_{k = 0}^{\infty} {(x - 2)}^{k} T o f i n d : V a l u e o f x S o l u t i o n : \sum_{k = 0}^{\infty} {(x - 2)}^{k} R a t i o t e s t a_{k} = {(x - 2)}^{k} a_{k + 1} = {(x - 2)}^{k + 1} L = \lim_{k \to \infty} |\frac{a_{k + 1}}{a_{k}}| T h i s s e r i e s w i l l b e c o n v e g e n t o f L < 1 \lim_{k \to \infty} |\frac{a_{k + 1}}{a_{k}}| < 1 \lim_{k \to \infty} |\frac{{(x - 2)}^{k + 1}}{{(x - 2)}^{k}}| < 1 \lim_{k \to \infty} |\frac{{(x - 2)}^{k} (x - 2)}{{(x - 2)}^{k}}| < 1 |x - 2| < 1 - 1 < x - 2 < 1 a d d 2 e a c h s i d e - 1 + 2 < x - 2 + 2 < 1 + 2 - 1 < x < 3$