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.
 A: 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 $c$. 
A: $\begin{array}\\
x_{n+1}-x_n
&=\sqrt{x_n+c}-x_n\\
&=(\sqrt{x_n+c}-x_n)\dfrac{\sqrt{x_n+c}+x_n}{\sqrt{x_n+c}+x_n}\\
&=\dfrac{x_n+c-x_n^2}{\sqrt{x_n+c}+x_n}\\
\end{array}
$
If 
$f(x) = x^2-x-c$,
$f'(x) = 2x-1$,
so $f'(x) > 0$
for $x > \frac12$.
The roots of $f(x)$
are
$x
=\dfrac{1\pm\sqrt{1+4c}}{2}
$,
so the positive root
$x_c$ satisfies
$1 < x_c
\lt 1+c$. 
If
$\frac12 < x_n < x_c$,
then
$x_n^2-x_n-c < 0$
so
$x_{n+1} > x_n$.
If
$ x_n > x_c$,
then
$x_n^2-x_n-c > 0$
so
$x_{n+1} < x_n$.
Similarly,
$\begin{array}\\
x_{n+1}-x_c
&=\sqrt{x_n+c}-x_c\\
&=(\sqrt{x_n+c}-x_c)\dfrac{\sqrt{x_n+c}+x_c}{\sqrt{x_n+c}+x_c}\\
&=\dfrac{x_n+c-x_c^2}{\sqrt{x_n+c}+x_c}\\
&=\dfrac{x_n+c-(x_c+c)}{\sqrt{x_n+c}+x_c}
\qquad\text{since } x^c_2 = x_c+c\\
&=\dfrac{x_n-x_c}{\sqrt{x_n+c}+x_c}\\
\end{array}
$
Therefore
$x_{n+1}-x_c$
has the same sign
and is smaller in absolute value
than
$x_n-x_c$.
Therefore
$x_n \to x_c$
since
$\sqrt{x_n+c}+x_c
\gt 1+c$.
