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Sum of geometric series

  1. Apr 28, 2014 #1
    For the question, shouldn't the sum be a(1/1-r) since we know lrl < 1 then that rn → 0 as n → ∞? I just don't quite understand why they wrote the sum is a(r/1-r). Is there a specific reason they did this? This is just a regular geometric series right? Is there any difference since the sum starts at n = 1 instead of n =0?
     

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  3. Apr 28, 2014 #2

    LCKurtz

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    Isn't the first term, which you are calling ##a##, equal to ##r## in that sum?
     
  4. Apr 28, 2014 #3
    Isn't ## a = \frac {1}{4}## and ##r = \frac{2}{e}##?
     
  5. Apr 28, 2014 #4
    Also, for this question, should the interval of convergence be ## [3,3]##? Why exactly did they say ##(-1,1)##?
     

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  6. Apr 28, 2014 #5

    Mark44

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    No. a = 1/(2e)

    Added: I'm talking about the first series in the image.
     
    Last edited: Apr 28, 2014
  7. Apr 28, 2014 #6

    Mark44

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    When you write fractions as text on a single line, you need to be careful about parentheses. 1/1-r means (1/1) - r, and not 1/(1 - r) as you intended.

    You have it below, as well.
     
  8. Apr 28, 2014 #7

    LCKurtz

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    In the picture where he has written ##\frac 1 4\sum_{k=1}^\infty r^k## the 1/4 is factored out and the remaining sum has ##a=r=\frac 2 e##.
     
  9. Apr 28, 2014 #8

    Ray Vickson

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    Of course there is a difference! If you start the sum at ##n = 0## the first term is ##a##; if you start it at ##n = 1## the first term is ##ar##. The two sums differ by ##a##.

    Anyway, they didn't write ##a(r/1-r) = ar - ar = 0##; they wrote ##ar/(1-r)##, where parentheses make all the difference in the world.
     
  10. Apr 29, 2014 #9

    Fredrik

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    The sum (##\sum_k r^k##) is 1/(1-r) when the summation starts at 0. If it starts at 1, you can do this: $$\sum_{k=1}^\infty r^k =r\sum_{k=1}^\infty r^{k-1} =r\sum_{k=0}^\infty r^k.$$
     
  11. Apr 29, 2014 #10
    So if it is instead ## ∑_{n=3}^∞r^n## then this equals ##∑_{n=0}^∞r^{n+3} = r^3∑_{n=0}^∞r^{n}## so then the sum is ##\frac{r^3}{1-r}## correct?
     
  12. Apr 29, 2014 #11

    LCKurtz

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    Yes. This agrees with the formula ##\frac a {1-r}## since the first term, ##a##, in ##\sum_{n=0}^∞r^{n+3}## is ##r^3##, as is the first term in ##\sum_{n=3}^∞r^n##. It doesn't matter how you write it.
     
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