How can I solve this infinite sum? I calculated (with the help of Maple) that the following infinite sum is equal to the fraction on the right side.
$$
\sum_{i=1}^\infty \frac{i}{\vartheta^{i}}=\frac{\vartheta}{(\vartheta-1)^2}
$$
However I don't understand how to derive it correctly. I've tried numerous approaches but none of them have worked out so far. Could someone please give me a hint on how to evaluate the infinite sum above and understand the derivation?
Thanks. :)
 A: Several good methods have been suggested. Here's one more. $$\eqalign{\sum{i\over\theta^i}&={1\over\theta}+{2\over\theta^2}+{3\over\theta^3}+{4\over\theta^4}+\cdots\cr&={1\over\theta}+{1\over\theta^2}+{1\over\theta^3}+{1\over\theta^4}+\cdots\cr&\qquad+{1\over\theta^2}+{1\over\theta^3}+{1\over\theta^4}+\cdots\cr&\qquad\qquad+{1\over\theta^3}+{1\over\theta^4}+\cdots\cr&\qquad\qquad\qquad+{1\over\theta^4}+\cdots\cr&={1/\theta\over1-(1/\theta)}+{1/\theta^2\over1-(1/\theta)}+{1/\theta^3\over1-(1/\theta)}+{1/\theta^4\over1-(1/\theta)}+\cdots\cr}$$ which is a geometric series which you can sum to get the answer. 
A: let $$S=\sum_{i=1}^\infty\frac{i}{\theta^i}, (\theta>1),$$
then
$$
\begin{align}
S-\frac{1}{\theta}S&=\sum_{i=1}^\infty\frac{i}{\theta^i}-\sum_{i=1}^\infty\frac{i}{\theta^{i+1}}\\
&=\sum_{i=1}^\infty\frac{i}{\theta^i}-\sum_{i=2}^\infty\frac{i-1}{\theta^{i}}\\
&=\frac{1}{\theta}+\sum_{i=2}^{\infty}\frac{1}{\theta^i}\\
&=\frac{1}{\theta}+\frac{1}{\theta^2-\theta},
\end{align}
$$
which yields
$$S=\frac{\theta}{(\theta-1)^2}$$
A: Hint: Consider the expectation (first moment) of the geometric distribution. Specifically, letting $1 - p = 1/\vartheta $ in the derivation of the formula ${\rm E}(Y)=(1-p)/p$ (in that link) gives exactly what you are looking for.
Elaborating. It is shown in the Wikipedia link how to derive the equality
$$
\sum\limits_{k = 0}^\infty  {(1 - p)^k pk}  = \frac{{1 - p}}{p}, \;\; 0 < p \leq 1.
$$
Letting $1-p = 1/\vartheta $, so that $p=(\vartheta - 1)/\vartheta$, this gives
$$
\sum\limits_{k = 1}^\infty  {k\frac{1}{{\vartheta ^k }}}  = \frac{{1 - p}}{{p^2 }} = \frac{1}{\vartheta }\frac{{\vartheta ^2 }}{{(\vartheta  - 1)^2 }} = \frac{\vartheta }{{(\vartheta  - 1)^2 }}.
$$
That is,
$$
\sum\limits_{i = 1}^\infty  {\frac{i}{{\vartheta ^i }}}  = \frac{\vartheta }{{(\vartheta  - 1)^2 }}.
$$
