How can I evaluate this infinite exponential series using a method I can learn?

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SUMMARY

The infinite series \(\sum_{n=0}^\infty n e^{-n\lambda}\) can be evaluated using differentiation techniques. The series can be transformed into a geometric series, where \(\sum_{n=0}^{\infty} e^{-n\lambda} = \frac{e^{\lambda}}{e^{\lambda}-1}\). By differentiating the geometric series with respect to \(\lambda\) and applying the appropriate transformations, the result is \(\frac{e^{\lambda}}{(e^\lambda-1)^2}\), as confirmed by Wolfram Alpha.

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ShayanJ
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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 want to learn it. Any ideas how to do it?
Thanks
 
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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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