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

srodriguex
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  • you can approximate it by an Gaussian integral, which is (relatively) easy, because solutions are generally known. – Ragnar Dec 27 '13 at 20:23
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    hint: set $u:=\dfrac 1{\theta}$ and compute $\displaystyle \frac d{du} \sum_{x=0}^\infty e^{-x,u}$. – Raymond Manzoni Dec 27 '13 at 20:24
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    The duplicate question that this is linked to is really impossible to find, because it has no keywords in the title that actually describe it. This question is much better, and has a better answer. – Praveen Sep 29 '16 at 21:41

2 Answers2

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

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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.

Matheman
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