What is the closed form for $ \sum_{n=1}^{\infty}n\theta^n, 0 \lt \theta \lt 1$? I have the following series:
$$ \sum_{n=1}^{\infty}n\theta^n, 0 \lt \theta \lt 1$$
I've seen lots of resources saying this is convergent. And using convergence tests to prove that. 
Question:

But is there any formula which gives the summation in closed form?  

 A: I guess the last two equations of this answer, (6) and (7), pack the real punch.  The rest is just . . .  commentary.
Working off the comment of amsmath, for $0 < \theta < 1$ we have
$\sum_1^\infty n \theta^n = \theta \sum_1^\infty n\theta^{n - 1} = \theta \dfrac{d}{d\theta}\sum_1^\infty \theta^n, \tag 1$
and
$\sum_0^\infty \theta^n = \dfrac{1}{1 - \theta}, \tag 2$
so
$\sum_1^\infty \theta^n = \dfrac{1}{1 - \theta} - 1 = (1 - \theta)^{-1} - 1, \tag 3$
whence
$\dfrac{d}{d\theta}\sum_1^\infty \theta^n = \dfrac{d}{d\theta}( (1 - \theta)^{-1} - 1) = (1 - \theta)^{-2}, \tag 4$
thus
$\sum_1^\infty n \theta^n = \dfrac{\theta}{(1 - \theta)^2}, \tag 5$
a closed-form expression.
We can in fact see how the factors $n$ arise if we look at the square of $(1 - \theta)^{-1}$:
$(1 - \theta)^{-2} = ((1 - \theta)^{-1})^2 = (\sum_0^\infty \theta^n)^2 = \sum_{n = 0}^\infty \sum_{i = 0}^n \theta^i \theta^{n -i} = \sum_0^\infty (n + 1) \theta^n, \tag 6$
whence
$\dfrac{\theta}{(1 - \theta)^2} = \theta \sum_0^\infty (n + 1) \theta^n =  \sum_0^\infty (n + 1) \theta^{n + 1} = \sum_1^\infty n\theta^n; \tag 7$
(6)-(7) seems to be another derivation of the desired result.
