The problem states:
Find the expectation $f(x; θ) = θ(1 − θ)^{x-1}$ when x = 1, 2, 3.... and 0<θ<1
I know I have to evaluate $$(x) = \sum_{x=1}^\infty xθ(1 − θ)^{x-1}$$
It sums to $$1/θ$$ but I don't understand why. What rule am I missing here? Sorry for formatting issues, I am learning
$$\sum_{x=1}^\infty x\theta(1-\theta)^{x-1}=-\theta\frac{d}{d\theta}\left(\sum_{x=1}^\infty (1-\theta)^x\right)=-\theta\frac d{d\theta}\left(\frac{1-\theta}{1-(1-\theta)}\right)=-\theta\frac d{d\theta}\left(\frac1\theta-1\right)=\frac1\theta.$$
See also: Geometric Series