For which values of $x$ does the sequence not converge? I have a wee problem. 
$$x_{i+1} = \frac{2}{18} (2x_i+1)^2$$
So - for what values of $x_1$ does the sequence not converge? 
Now, some pesky classmate has borrowed my notes, so I ran out of ideas pretty quickly. I'm sure the solution is quite obvious, but for the life of me, I couldn't make this blooming sequence diverge. 
Any ideas, hints, or suggestions could be greatly appreciated. 
 A: By the ratio test, we have
$$\frac {\frac 19(2x_i+1)^2}{x_i}=\frac {4x_i^2+4x_i+1}{9x_i}= \frac 49(x_i+1)+\frac 1{9x_i}\le 1$$
For the ratio test to succeed, we need $\frac {x_{i+1}}{x_i}\le 1$.  Multiplying by $\frac 94$ through while comparing to $1$, we have $(x_i+1)+\frac 1{4x_i}\le \frac 94$ and  $x_i+\frac 1{4x_i}\le \frac 54$ which means that $x_i^2+\frac14\le \frac {5x_i}4$ or $x_i^2-\frac 54x_i+\frac14\le 0$ which is an easy-to-solve quadratic:
$$x_i\le\frac{\frac 54\pm\sqrt{\frac {25}{16}-1}}{2}=\frac 58\pm\sqrt{\frac 9{64}}=\frac 58\pm\frac38$$
So the set of solutions to this series is in the interval $x_1\in[\frac 14,1]$.  Since for all $i\ge 2$, $x_i\ge 0$, we can check $x_1\lt 0$ for the values $x_2\in[\frac 14, 1]$:
$$\frac 14\le \frac {(2x_1+1)^2}9\le 1$$
or
$$-\frac 32\ge 2x+1\ge -3\\-\frac 54\ge x\ge-2$$
So the full set of solutions is $x_1\in [\frac 14,1]\cup[-2,-\frac 54]$.  Wait, trying $x_1=0$ yields $x_2=\frac 19,x_3=\frac{121}{729},\dots x_i\to \frac 14$.  So we really have the full set of solutions $x_1\in[-2,1]$.
