Motivation for forming a specific bound for a recursively defined sequence. Often, it is not easy to observe a specific bound for a recursively defined sequence.
The simplest method would be to write a finite number of terms and observe the pattern.
However, in many cases, this approach is not useful.
For example,
Suppose $c>0$ and $0<a<\frac{1}{c}$. Define the sequence $(x_{n})$ as follows,
$$x_{0}=a,\text{  } x_{n}=x_{n-1}(2-cx_{n-1}) \text{    for   }n=1,2,...$$
The original question is to prove that the sequence converges to $\frac{1}{c}$.
It was said that the first step is to show that $0<x_{n}<\frac{1}{c}$ for all $n=0,1,2,...$
My question:
How can we tell that the inequality holds without beforehand knowing it is true?
What is the motivation? 
Any useful tricks i can use for this situation?
Any kind person can help me? i will be greatly grateful. Thanks
 A: Personally, I check a calculator to see if the sequence is monotone (increasing or decreasing).  This is not a proof of course, but it does give you a good idea of what you want to prove.
Then, I assume it exists if it looks like it does or if it doesn't, then I prove it doesn't.  At this step, I substitute $L=\lim_{n\to\infty} x_n$ to get
$$L=L(2-cL)$$
Solving this with the assumption that $L>0$ gives
$$L=\frac1c$$
Note that assuming $L>0$ is not a big problem since we can clearly see we should have $L>0$ in the calculator.  Likewise, we can see it should be monotone increasing.  Thus, putting two and two together, we want to prove the following:
$$\text{Verify this is monotone:}\quad x_{n+1}>x_n(>0)$$
$$\text{Show that it is bounded above:}\quad x_n<~?$$
Now, what should be our upper bound you ask?  For problems like this, it almost always works beautifully to set the upper bound we want to prove to be the limit itself.
$$x_n<\frac1c\tag1$$
Now, let's prove:
$$\begin{align}x_{n+1}&=x_n(2-cx_n)\\(1)~&>x_n\left(2-c\frac1c\right)\\&=x_n(2-1)\\&=x_n\end{align}$$
where $(1)$ uses our assumed upper bound $x_n<\frac1c$.  Now we prove the upper bound without any assumptions to avoid paradoxes:
First, notice that
$$\left(\sqrt cx_n-\frac1{\sqrt c}\right)^2=cx_n^2-2x_n+\frac1c\tag2$$
It thus follows that
$$\begin{align}x_{n+1}&=x_n(2-cx_n)\\&=2x_n-cx_n^2\\(2)~&=\frac1c-\left(\sqrt cx_n-\frac1{\sqrt c}\right)^2\\&<\frac1c\end{align}$$
And we are done!
