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Infinite exponential series

  1. Mar 20, 2015 #1


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    Gold Member

    I've just encountered the infinite series ## \sum_{n=0}^\infty n e^{-n\lambda} ##. I know that in general its not possible to evaluate an infinite series but because wolframalpha.com could evaluate it(which gave the result ## \frac{e^{\lambda}}{(e^\lambda-1)^2} ##), it seems to me that this one has can be solved by a method that I can learn. So I wanna learn it. Any ideas how to do it?
  2. jcsd
  3. Mar 20, 2015 #2


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    Science Advisor

    Here are the elements of a solution. Some of the justifications must be verified, but here goes:
    [itex]ne^{-n\lambda}= \frac{d}{d\lambda}(-e^{-n\lambda}) [/itex]. The series [itex]e^{-n\lambda} [/itex]is geometric, with ratio [itex]e^{-\lambda} [/itex], and thus [itex]\sum_{n=0}^{\infty}e^{-n\lambda}=\frac{1}{1-e^{-\lambda}}=\frac{e^{\lambda}}{e^{\lambda}-1} [/itex]. Change the sign and derive with respect to λ and you are there.
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