Find all possible values of $a, p$ for which $x_{n+1} = x_n^2 + (1-2p)x_n + p^2$ converges, given $x_1 = a$ and $n\in\Bbb N$ 
Given a recurrence relation 
  $$
x_{n+1} = x_n^2 + (1-2p)x_n + p^2\\
x_1 = a\\
n\in\Bbb N
$$
  find all values of $a, p \in\Bbb R$ for which $x_n$ converges.

First of all I tried to assume that the limit indeed exists. Then it must match a fixed point of the following equation:
$$
x = x^2 + (1-2p)x + p^2
$$
Which solving for $x$ gives:
$$
(x-p)^2 = 0 \iff x = p
$$
By this if the recurrence converges then it must follow that:
$$
\lim_{n\to\infty}x_n = p
$$
Lets now take a closer look at the recurrence. I've shown that it must describe a monotonically increasing sequence, namely:
$$
x_{n+1} = x_n^2 + (1-2p)x_n + p^2 \\
x_{n+1} = x_n^2 - 2px_n + p^2 + x_n \\
x_{n+1} = (x_n - p)^2 + x_n \\
x_{n+1} - x_n = (x_n - p)^2 > 0 \implies \boxed{x_{n+1} > x_n}
$$
By this the sequence in monotonically increasing no matter what initial conditions are given. 
At this point I'm lost. How do one deduce the constraints for $a, p$ for $x_n$ to be convergent. Since the sequence is increasing I guess the problem may be reduced to "find the values of $a, p$ for which the sequence is bounded". Then the result should follow by monotone convergence theorem. The answer section suggests that:
$$
0 \le p - a \le 1
$$
Which seems true based on the Cobweb plot. But a plot in not a formal proof. Unfortunately I was not able to infer that result. What is a proper way to finish the problem? 
Please note this problem is given in the limits section. So even derivatives are not available.
 A: Instead of viewing it as a map on $x_n$, try examining what happens to $x_n-p$.
From the second to last line of your display, it's easy to work out that
$(x_{n+1}-p) = (x_n-p)^2 + (x_n-p).$
So the sequence $b_n=x_n-p$ (which converges exactly when $x_n$ does) is iterating under them map $t\mapsto t^2+t$.  Now you have one function to understand not a class.  It sounds like you have the tools necessary to verify this converges if the initial value $b_1$ is between -1 and 0.
A: A first simplification is to define $y_n=x_n-p$. Then as you already proved
$$y_{n+1} -y_n=y_n^2\\
y_1 = a-p\\
n\in\Bbb N$$
the problem is now to find conditions on $y_1$ such that $y_n \to 0$, i.e to study the recurrence relation $y_{n+1}=f(y_n)$ with the fixed (independent of $a$ and $p$) function $f(x)=x+x^2$.

As $y_{n+1} -y_n=y_n^2$ the sequence is increasing, and the only possible limits are $-1$ and $0$ so


*

*if $y_1 >0$ then for all $n$, $y_n>0$ and the series does not converge.

*if $y_1 <-1$ then $y_2>0$ and once more for all $n$, $y_n>0$ and the series does not converge.

*if $-1 \leq y_1 \leq 0$ you can show by recurrence that for all $n$ $-1 \leq y_n \leq 0$, so $y_n$ is a bounded monotonous function and thus converges.


You finally obtain the condition
$$-1 \leq y_1 \leq 0$$
i.e
$$-1 \leq a-p \leq 0$$
