How to compute infinite series $\sum_{n=0}^{\infty} ne^{-n}$

Question

I'm trying to compute the infinite series $\sum_{n=0}^{\infty} ne^{-n}$.

I know the answer is $e/(e-1)^2$, but I don't understand how to find this result.

Thanks for the help!

Answer 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 $x = 1$.

Answer 2

Another approch: $s=\sum_{n=0}^\infty e^{-nx}$ So,$s=e^{-1}+2e^{-2}+...+$

Now if you consider $(s-e^{-1}s)$ you will get an infinite gp series. Then solving you can find $s.$