What range of values for $q\in[0, 1]$ imply the convergence of $x_{n+1} = 1+qx_n^2$ 
Let:
  $$
x_1  = a\\
0<a<1\\
x_{n+1} = 1+qx_n^2\\
n\in \Bbb N
$$
  For what values of $q\in[0, 1]$ the sequence $\{x_n\}$ is convergent?

I've started with the following. We are given that $a\in(0,1)$, then:
$$
x_1 = a\\
x_2 = 1+qa^2
$$
Obviously since $a<1$:
$$
x_1 < x_2
$$
Suppose $\forall n\in\Bbb N: x_{n+1}>x_n$:
$$
x_{n+1} > x_n \\
x_{n+1}^2 > x_n^2 \\ 
qx_{n+1}^2 > qx_n^2 \\
1+qx_{n+1}^2 > 1+qx_n^2 \\
x_{n+2} > x_{n+1}
$$
The induction shows $x_n$ is monotonically increasing. A monotonic sequence is convergent when it's bounded. Suppose $x_n$ is convergent. Let's try to find the fixed point of the recurrence:
$$
x = 1 + qx^2\\
qx^2 - x + 1 = 0 \\
x = \frac{1\pm \sqrt{1-4q}}{2q}
$$
For a fixed point to exist (at least in $\Bbb R$) we need:
$$
1-4q \ge 0 \\
q\le {1\over 4}
$$
Or given initial conditions for $q$:
$$
0\le q\le {1\over 4}
$$
However the solution above is missing upper bound for $x_n$ which I was not able to find. What would be the way to finish the problem? (In the answer section $q$ is indeed given in $q\in\left[0, {1\over 4}\right]$, but i still need to justify that)
 A: *

*$(x_n)$converges for $0\le q\le \frac{1}{4}$:  Let $\alpha=\frac{1+\sqrt{1-4q}}{2q}$ be the larger root of the equation $qt^2-t+1=0$. Since $\frac{1}{\alpha}$ is the other root, we find that $\alpha \ge 1\ge x_0$. We also find that
$$
\alpha \ge x_n \implies \alpha =q\alpha^2 +1\ge qx_n^2 +1=x_{n+1}.
$$ By induction, we have $\alpha \ge x_n$ for all $n$. Combining with the fact (in the OP) that $(x_n)$ is increasing, there exists
$$
l=\lim_{n\to \infty} x_n \le \alpha
$$ by monotone convergence theorem.


*$(x_n)$ diverges (to $\infty$) for $q>\frac{1}{4}$: We find that
$$
qt^2 -t+1 =q\left(t-\frac{1}{2q}\right)^2+1-\frac{1}{4q}\ge 1-\frac{1}{4q}=c>0.
$$ This implies
$$
x_{n+1} =qx_n^2+1 \ge x_n +c,\quad\forall n.
$$ Therefore, by induction we have
$$
x_n \ge x_0 +nc \stackrel{n\to\infty}\longrightarrow \infty.
$$ Summing up, $(x_n)$ converges if and only if $q\in [0,\frac{1}{4}].$
A: Assume $0\le q\le1/4$ and let
$$
x^*=\frac{1-\sqrt{1-4\,q}}{2\,q}=\frac{2}{1+\sqrt{1-4\,q}}.
$$
$x^*$ is a fixed point and $x^*>1$ (in particular, $a<x^*$.) We have also that
$$
x<1+q\,x^2,\quad 0<x<x^*.
$$
Since $0<a<x^*$, it follows that $0<x_n<x^*$ for all $n\in\Bbb N$.
