A solution to a nonlinear recurrence equation I am interested in finding a solution to the following recurrence relationship:
$$
F_{n+1} = \sqrt{ A + B/F_n }
$$
OR equivalently, the equation: 
$$
S_{n+1} = C + \frac{E}{\sqrt{S_n}}
$$
where $A,B, C, E$ are real positive numbers.
 I would appreciate to have a solution for any value of n, or at least to know if these sequences are bounded, or if they converge at all. The starting points are nonzero positive reals.

Many thanks in advance.

Thanks for those who help. I would like to add to this question, the possibility of having A and B being independently negative. Also, Does taking the absolute value inside the square root force convergence?
I mean: is the relation $$ F_{n+1} = \sqrt{ |A + B/F_n | } $$ better ( i.e. more guaranteed, i.e. wider range of possibilities ) to converge than the one above for any real value for A and B (positive or negative)? 
Since F(0) is nonzero positive, we cannot be sure that F(n) is not zero now we are having negative values for A and/or B. So, any guess of a starting point to guarantee convergence at this case is also well appreciated.

Note:  I was reading the work of Rabinovic http://www.math.tamu.edu/~berko/papers/pdf/jmpRBH96.pdf ) , and I think the theorem reported therein can be a great basis for solving any nonlinear recursion if it could be written as a polynomial. A strategy that comes in my mind is: To expand the relation using Taylor series, and then write the general polynomial term as a function of F(n) . Then use the theorem reported in the that paper,
Is this idea sound? 

Sorry for long blathering
In this case, a sketch is  very worthy.



It is clear how the recurrence is graphed, and how it is developing out.

We consider only the domain $0&lt;x$.
In this function has an horizontal asymptote  $y_{\infty}= \sqrt{A}$, and remains above that.
The derivative 
$$
y'(x) = {d \over {dx}}\sqrt {A + B/x}  =  - \;{B \over {2x^{\,2} \sqrt {A + B/x} }}
$$
is always negative if $0&lt;B$.
The second derivative is always positive, for positive $A$ and $B$.

Then, since in the recursion remains $A&lt;x$, if the absolute value of the 
derivative at $x=\sqrt{A}$ is less than $1$ 
$$
\left| {y'(\sqrt A )} \right| = \;{B \over {2A\sqrt {A + B/\sqrt A } }} &lt; 1
$$
the Priciple of Contraction guarantees that the recurrence 
is convergent to $x=y(x)$.
HINT

Consider the sequence $(F_n)_{n \ge 0}$, given by the recurrence relation
$$
F_{n+1} = \sqrt{A+ \frac{B}{F_n}}, \quad F_0, A, B \in \mathbb{R}^+.
$$
If this sequence converges to some limit $F$, then $F$ must satisfy
$$
F = \sqrt{A + B/F}
$$
so you must solve $F^3 - AF - B = 0$, which has exact solutions using Cardano's formula. At least one of them is guaranteed to be real.



The second recurrence will similarly lead to $$(S-C)^2 = \frac{E^2}{S},$$ which is a cubic as well...