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Divergence of the geometric Series at r=1

  1. Apr 29, 2012 #1
    Now for the proof of convergence/divergence of the geometric series we first deduce the Nth partial sum which is given by:

    [tex] \frac{r(1-r^n)}{1-r} [/tex]

    Now for 0<r<1 this become [tex] \frac{1}{1-r} [/tex] which clearly converges by AOL

    At r>1 it's similarly obvious why it diverges.

    But at r=1, I'm a bit confused. The text say's that it diverges but when I try to work it out I get:

    [tex] 1^n -> 1 [/tex]

    [tex] S_n = \frac{C*0}{1-1} [/tex]

    which measn S_n = 0 which is confusing as it seems to indicate convergence ?

    Am I suppose to interpret 0/0 as implying divergence ?
     
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  3. Apr 29, 2012 #2

    sharks

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    For a geometric series, |r|<1 for convergence. If r has any other value, including 1 and -1, the series diverges.

    If you find the limit at r=1, you will get a/(1-1)=a/0=∞, hence the series diverges.
     
  4. Apr 29, 2012 #3

    Ray Vickson

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    You are supposed to avoid 0/0 like the plague. The point is that in the LIMIT as r --> 1, the sum r/(1-r) --> infinity. Of course, you can also write down the sum directly when r = 1, to see that S = 1 + 1 + 1 + 1 + .... does not have a finite value.

    RGV
     
  5. Apr 29, 2012 #4

    HallsofIvy

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    What you should learn from this is "Don't just use formulas thoughtlessly- think!

    If r= 1, then the sum is [itex]\sum_{i=0}^\infty a(1)^n= a+ a+ a+ \cdot\cdot\cdot[/itex]. The nth partia series is na so that clearly does not converge.

    If r= -1, then the sum is [itex]\sum_{i= 0}^\infty a(-1)^n= a- a+ a- a+ \cdot\cdot\cdot[/itex]. The nth partial series is a if n is odd, 0 if n is even so, again, that does not converge.
     
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