# Homework Help: Calculus II - Infinite Series

1. Sep 18, 2011

### GreenPrint

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!

2. Sep 18, 2011

### Ray Vickson

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

3. Sep 18, 2011

### GreenPrint

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?

4. Sep 18, 2011

### romsofia

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).

5. Sep 18, 2011

### GreenPrint

hm interesting thanks

6. Sep 18, 2011

### romsofia

Sorry that I didn't answer your second question, if $${n<1}$$ it will ALWAYS converge, if $${n>1}$$ it will ALWAYS diverge! If $${n<=1}$$ 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.