Convergence of a sum for which x?

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Homework Help Overview

The discussion revolves around the convergence of the infinite series (1/n) * (x^n), where x is a real number. Participants are exploring the conditions under which the series converges, converges absolutely, diverges to positive infinity, or does not converge at all.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the convergence behavior of the series for various values of x, including specific cases such as x = 1, x = -1, and x > 1. Questions arise regarding the behavior of the series for values of x between -1 and 1.

Discussion Status

Some participants have reached conclusions about convergence for |x| < 1 and divergence for |x| > 1, while others are still questioning the behavior at the endpoints x = 1 and x = -1. There is an ongoing exploration of the nature of the series at these critical points, with some guidance provided regarding the alternating series at x = -1.

Contextual Notes

Participants are navigating the implications of the ratio test and the characteristics of alternating series, with some uncertainty remaining about the convergence at specific boundary values.

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Homework Statement


Consider the infinite series (1/n) * (xn) where x is a real noumber. Find all numbers x such that
i) the series converges,
ii) series converges absolutely
iii) diverges to + infinity
iiii) does not converge.




2. The attempt at a solution
For this, i know it does not converge for x = 1 and for x<-1 , and it does converge absolutely for x = -1, i also know it will diverge to [tex]\infty[/tex] for x>1 but i am very unsure as to what happens for x between 1 and - 1 ?

Thanks a lot
 
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What does the ratio test tell you?
 
Ok, I see, so now i have that it converges for |x|<1 and the same for absolute convergence.
It does not converge at x=1 , x=-1 pr x<-1
and it diverges to infinity for x>1.
It this closer?
 
stukbv said:
Ok, I see, so now i have that it converges for |x|<1 and the same for absolute convergence.
It does not converge at x=1 , x=-1 pr x<-1
and it diverges to infinity for x>1.
It this closer?

Closer. It converges for |x|<1 and diverges for |x|>1. The only points you have to worry about are x=1 and x=(-1). I'll agree it diverges at x=1. Can you say why? I don't agree that it diverges at x=(-1).
 
For x= -1, this is an alternating series and 1/n goes to 0.
 
Oh yes, of course!
Thanks!
 

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