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How to derive [itex]\lim_{n\rightarrow ∞} \frac{1-r^{N+1}}{1-r}[/itex]

  1. May 29, 2013 #1
    How is [itex]\lim_{n\rightarrow ∞} \frac{1-r^{N+1}}{1-r}[/itex] derived?

    Is it directly related to [itex]S_{n}=\frac{a(1-r^{n})}{1-r}[/itex]?

    If so, how does the [itex]r^{n}[/itex] become [itex]r^{n+1}[/itex]

    and how does the a disappear?

    I've seen how to derive [itex]S_{n}=\frac{a(1-r^{n})}{1-r}[/itex], but have never come across [itex]\lim_{n\rightarrow ∞} \frac{1-r^{N+1}}{1-r}[/itex] until I just came across a brief mention of it in a video.

    Any suggestions on where I could find a derivation/proof for it?
     
  2. jcsd
  3. May 29, 2013 #2

    Office_Shredder

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    You have an N in the fraction but you're taking the limit for n. I'm assuming these are supposed to be the same thing.

    There are several possibilities:

    1) r > 1. Then rn gets really big as n goes to infinity, so this limit diverges
    2) r = 1. Then the numerator and denominator are both zero and you shouldn't have written this down to begin with
    3) r < 1. Then rn gets really small as n goes to infinity, so the limit converges and it converges to [tex] \frac{1}{1-r} [/tex]

    Are you looking for a rigorous proof of this, or is a description of the behavior like this sufficient?
     
  4. May 30, 2013 #3
    I was looking for a proof of where [itex]\frac{1-r^{N+1}}{1-r}[/itex] comes from and what it means. I think I found a proof for it, but still trying to absorb it.
     
  5. May 30, 2013 #4

    HallsofIvy

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    change the numbering: Let j= n-1 and n= j+ 1 so [itex]r^n= r^{j+1}[/itex]. Of course, as n goes to 0, so does j. And it doesn't matter what "name" you give the parameter so just call it "n" now instead of "j".

    Take a= 1, of course.

     
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