Stuck on a recursively defined converging sequence problem. I am given a simple quadratic equation
$$x^2-x-c=0, x>0, c>0$$
and then we define a sequence $\{x_n\}$ with $x_1>0$ fixed and then, if $n$ is an index for which $x_n$ has been defined, we define
$$x_{n+1}=\sqrt{c+x_n}$$.
With that I am asked to prove that $\{x_n\}$ converges monotonically to the solution of the polynomial.
I've done quite a bit of scratch work. Obviously we can solve the quadratic and the positive solution is $\frac{1+\sqrt{5}}{2}$. I have an inkling that the equation is decreasing and so I tried working with $x_{n+1}-x_{n+2}$ to show that the difference is positive but I didn't come up with anything useful.
I did realize that if I simply write out the limit we see that
$$\lim_{n\rightarrow\infty}[(\sqrt{c+x_{n+1}})^2)-\sqrt{c+x_{n+1}}-c]=\lim_{n\rightarrow\infty}[x_{n+1}-\sqrt{c+x_{n+1}}]$$
So if we want this final limit to go to zero then all I really need is $x_n$ to be monotonically decreasing since it is clearly bounded below by zero since $c$ and $x_1$ were taken to be positive.

Hint
$$x_{n+1}^2-x_n=c$$
$$x_{n+2}^2-x_{n+1}=c$$
so,
$$x_{n+2}^2-x_{n+1}^2-(x_{n+1}-x_n)=0$$
$$(x_{n+2}-x_{n+1})(x_{n+2}+x_{n+1})=(x_{n+1}-x_n)$$
suppose that $x_{N+1}<x_{N}$ for some $N$. What can you conclude?
After that, you have to study the relation between $x_1$ and $x_2$, which will depends on $c$. 
