Infinite exponential series

1. Mar 20, 2015

ShayanJ

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?
Thanks

2. Mar 20, 2015

Svein

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

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