How do we determine the convergence at the endpoints for the series (x-1)^n/n^3?

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Homework Help Overview

The discussion revolves around determining the interval of convergence for the series \(\sum_{n=0}^{\infty} \frac{(x-1)^n}{n^3}\). Participants are exploring the implications of the root test and the necessity of testing endpoints in the context of convergence.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand why testing endpoints is necessary when the root test yields a limit of 1, which they believe indicates convergence. Other participants clarify that a limit of 1 does not guarantee convergence and provide examples to illustrate this point.

Discussion Status

Participants are actively questioning the assumptions behind the root test and discussing the conditions under which convergence can be determined. There is an acknowledgment that further testing is required at the endpoints, and some guidance has been provided regarding the implications of the root test.

Contextual Notes

There is some confusion regarding the interpretation of the root test, particularly in relation to the values of \(r\) and the conditions for convergence. Participants are also referencing specific cases where the root test is inconclusive.

GreenPrint
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Hi,

I don't really need help with a problem, just having some troubles understanding something.

Find the interval of convergence of the series

sigma[n=0,inf] (x-1)^n/n^3
by the root test I got that |x-1|<1 and that 0<x<2
I than have to plug in these values (0 and 2) to see if the series converges or diverges at these endpoints... I however am confused by this. If we set x = 0 or x = 2 wouldn't the root test give us 1 and the root test states that if the limit equals one than the series converges. So what's the need of testing the end points? Like I know we have to just am not really sure why because than the root test would give us 1... I never understood why we test the end points and just sort of did so just because I was told we had to and was wondering if someone could explain this to me. I hope someone can clear up this confusion for me. Thanks for any help!
 
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GreenPrint said:
Hi,

I don't really need help with a problem, just having some troubles understanding something.

Find the interval of convergence of the series

sigma[n=0,inf] (x-1)^n/n^3
by the root test I got that |x-1|<1 and that 0<x<2
I than have to plug in these values (0 and 2) to see if the series converges or diverges at these endpoints... I however am confused by this. If we set x = 0 or x = 2 wouldn't the root test give us 1 and the root test states that if the limit equals one than the series converges. So what's the need of testing the end points? Like I know we have to just am not really sure why because than the root test would give us 1... I never understood why we test the end points and just sort of did so just because I was told we had to and was wondering if someone could explain this to me. I hope someone can clear up this confusion for me. Thanks for any help!

Root test = 1 does NOT guarantee convergence. Look at sum_{n=1}^infinity 1. Does it converge? Does it satisfy root test = 1? Look at sum r^n /n. Root test implies convergence for |r| <= 1. Does the series converge at r = 1? At r = -1?

RGV
 
Well like I forgot myself and had to look up on wikipedia http://en.wikipedia.org/wiki/Root_test and it says that if r=1 than the series diverges so I thought it was always |r| < 1 not |r| <= 1?
 
If n=1 when using the root test, it means your answer is inconclusive AKA use another test (that's what RGV is hinting at).
 
hm interesting thanks
 
Sorry that I didn't answer your second question, if [tex]{n<1}[/tex] it will ALWAYS converge, if [tex]{n>1}[/tex] it will ALWAYS diverge! If [tex]{n<=1}[/tex] then I'm going to guess that it'll converge, but I've never seen a series that has n<=1!

This is for ratio/root test.
 

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