# 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!

• Let your first sum be $S$. Would you mind evaluating $S - PS$? Commented May 13, 2016 at 22:08
• I've evaluated as $\frac{1}{1-P} - \frac{P}{1-P} = 1$, although I don't understand the significance of this result...(except that the result is certain?) Commented May 13, 2016 at 22:11

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