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Infinite sum of infinite negative series e^(-nt)

  1. Apr 15, 2012 #1
    Find the sum of
    e^(-nt) as n = 0,1,2,3,...

    Apparently it's 1/(1-e^-t)

    Can't use normal geometric series formula though as if a =1, r = e?
    Any ideas
    Thanks :)
     
  2. jcsd
  3. Apr 15, 2012 #2

    Curious3141

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    Homework Helper

    As long as t>0, you can treat this as the sum of an infinite geometric series.

    The common ratio is e^(-t), which has magnitude less than 1. First term is e^0 = 1.

    The sum is a/(1-r) = 1/(1-e^(-t)).
     
  4. Apr 15, 2012 #3
    of course when the condition of a=1 occurs it's just 1/(1-r). it was very easy then. doh :)
     
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