Find the interval of convergence

Click For Summary

Homework Help Overview

The discussion revolves around finding the interval of convergence for the power series ∑(x-2)n / 3n. Participants are examining the convergence criteria based on the ratio test and are comparing their findings with a provided answer from a textbook.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the ratio test, specifically calculating ρn and setting it to be less than or equal to one to determine convergence. There is confusion regarding the endpoints of the interval of convergence, particularly the discrepancy between their calculations and the textbook answer.

Discussion Status

The discussion is active, with participants clarifying their understanding of the interval versus radius of convergence. One participant acknowledges a mistake in their interpretation of the endpoints, leading to a productive exchange about the conditions under which endpoints are included or excluded.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific rules regarding the treatment of endpoints in convergence intervals.

acdurbin953
Messages
42
Reaction score
0

Homework Statement


Find the interval of convergence of the power series ∑(x-2)n / 3n

Homework Equations


ρn = |an+1| / |an|

The Attempt at a Solution


I got that ρn = | (x-2) / 3 |. I set my ρn ≤ 1, since this is when the series would be convergent. Manipulating that expression, I got that the interval of convergence is -5 ≤ x ≤ 5. The answer in the back of the book is -1 ≤ x ≤ 5. I am confused where the -1 comes from.
 
Physics news on Phys.org
I'm sorry, I realized after this post that I was getting interval of convergence mixed up with radius of convergence.
 
acdurbin953 said:

Homework Statement


Find the interval of convergence of the power series ∑(x-2)n / 3n

Homework Equations


ρn = |an+1| / |an|

The Attempt at a Solution


I got that ρn = | (x-2) / 3 |. I set my ρn ≤ 1, since this is when the series would be convergent. Manipulating that expression, I got that the interval of convergence is -5 ≤ x ≤ 5. The answer in the back of the book is -1 ≤ x ≤ 5. I am confused where the -1 comes from.

Be careful: the endpoints are generally not included, because they would correspond to a series of the form ##\sum 1^n## or ##\sum (-1)^n##. Are you sure the book did not say ##-1 < x < 5##?
 
Ray Vickson said:
Be careful: the endpoints are generally not included, because they would correspond to a series of the form ##\sum 1^n## or ##\sum (-1)^n##. Are you sure the book did not say ##-1 < x < 5##?

You're right, it does say -1 < x < 5. Thank you for that reminder!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K