Explanation of the Sum of an Infinite Series Equation I've been presented with the following infinite sum (where $P$ is the probability of an event, and $1-P$ is therefore the probability of it not occurring.
I was given the following equation as fact:
$$\sum_{i=1}^\infty(iP^{i-1}(1-P))=\frac{1}{1-p}$$
Now I know from my basic education the sum of an infinite series for probability $P$ is
$$ \sum_{i = 0}^\infty P^i = \frac{1}{1-P}$$
Assuming I haven't made a very basic mistake, these are therefore equivalent but I don't really see how. Normally, I would ask the person who proposed the first equation to justify their working, however, in this instance I can't (they have an obscure transform named after them and would take it very badly!).
My question is, how is the first equation evaluated to such a simple result, and therefore equivalent to the second.
Thanks!
 A: You know that 
$$\sum_0^\infty p^i=\frac{1}{1-p}$$
for every $p$ such that $|p|\lt 1$. Treat $p$ as a variable, and differentiate both sides with respect to $p$. We get
$$\sum_{i=0}^\infty ip^{i-1}=\frac{1}{(1-p)^2}.$$
The sum can be taken as starting at $i=1$. If we multiply both sides by $1-p$, we arrive at the result that you want to prove.
A: Apply algebra to the convergent series.
$$\begin{align}\sum_{i=1}^\infty ip^{i-1}(1-p)
~=~&p^0+\sum_{i=2}^\infty ip^{i-1}-\sum_{i=1}^\infty ip^i
\\~=~&p^0+\sum_{i=2}^\infty ip^{i-1}-\sum_{i=2}^\infty (i-1)p^{i-1}
\\~=~&p^0+\sum_{i=2}^\infty p^{i-1}
\\~=~&\sum_{i=0}^\infty p^i \end{align}$$
A: One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by making $n \to +\infty$ in $(2)$, using $|x|<1$, one gets 

$$
\sum_{i=0}^\infty i x^{i-1}=\frac1{(1-x)^2}. \tag3
$$

