How can I resolve $\sum_{x=0}^{\infty} xe^{-x/\theta}$?

Question

I stumble on this summation during an exercise. How can I resolve $\sum_{x=0}^{\infty} xe^{-x/\theta}$?

Answer 1

Here is a slightly more general strategy that can be adapted here: If $|r| < 1$, we have

$$\sum\limits_{x = 0}^{\infty} r^x = \frac{1}{1 - r}$$

Taking a derivative on both sides leads to

$$\sum\limits_{x = 1}^{\infty} x r^{x - 1} = \frac{1}{(1 - r)^2}$$

or by a change of indices,

$$\sum\limits_{x = 0}^{\infty} (x + 1) r^{x} = \frac{1}{(1 - r)^2}$$

So applying this last statement combined with the first,

$$\sum\limits_{x = 0}^{\infty} x r^x = \frac{1}{(1 - r)^2} - \frac{1}{1 - r} = \frac{r}{(1 - r)^2}$$

So choose $$r = e^{-1/\theta}$$

Answer 2

In addition to the general solution, you have to cut off those terms with $\theta <0$, because the general term $xe^{-\frac{x}{\theta}}$ diverges and the series cannot converge.