In the book "Contours, Convex Sets and Cellular Automata" (Andrei Toom), I've found the following:
[...] the probability that there is no percolation does not exceed
$\sum_{n=2}^{\infty}(n-1) \cdot 3^{2n-2}(1-\varepsilon )^{2n}$
For (1 - ε) small enough this sum is less than one and this is what we need. In fact, this sum equals
$\left ( \frac{x}{3(1-x)} \right )^2$ where $x = \left ( 3(1-\varepsilon ) \right )^2$.
I did not understand how he derived the summation to the other expression. Can someone give me a help?
If $x=\big(3(1-\epsilon)\big)^2$, then
$$\begin{align*} \sum_{n\ge 2}(n-1)3^{2n-2}(1-\epsilon)^{2n}&=\sum_{n\ge 2}(n-1)\big(3(1-\epsilon)\big)^{2(n-1)}(1-\epsilon)^2\\ &=\sum_{n\ge 2}(n-1)x^{n-1}(1-\epsilon)^2\\ &=\frac19\sum_{n\ge 2}(n-1)x^n\\ &=\frac{x^2}9\sum_{n\ge 2}(n-1)x^{n-2}\\ &=\frac{x^2}9\sum_{n\ge 1}nx^{n-1}\\ &=\frac{x^2}9\cdot\frac{d}{dx}\left(\sum_{n\ge 0}x^n\right)\\ &=\frac{x^2}9\cdot\frac{d}{dx}\left(\frac1{1-x}\right)\\ &=\frac{x^2}9\cdot\frac1{(1-x)^2}\\ &=\left(\frac{x}{3(1-x)}\right)^2\,. \end{align*}$$