How to compute infinite series $\sum_{n=0}^{\infty} ne^{-n}$
Solution 1:
You can differentiate the series $\sum_{n=0}^\infty e^{-nx} = \frac{1}{1-e^{-x}}$ to obtain $\frac{d}{dx} \sum_{n=0}^\infty e^{-nx} = -\sum_{n=0}^\infty n e^{-nx}$. Then just put $x = 1$.