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Calculus II - Infinite Series

  1. Sep 18, 2011 #1
    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. jcsd
  3. Sep 18, 2011 #2

    Ray Vickson

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    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
     
  4. Sep 18, 2011 #3
    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?
     
  5. Sep 18, 2011 #4
    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).
     
  6. Sep 18, 2011 #5
    hm interesting thanks
     
  7. Sep 18, 2011 #6
    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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