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. :)

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,
$$
\sum\limits_{i = 1}^\infty  {\frac{i}{{\vartheta ^i }}}  = \frac{\vartheta }{{(\vartheta  - 1)^2 }}.
$$
