I am trying to solve a nonlinear recursive relation that reads \begin{equation} \frac{1}{f(n+1)} = \frac{1}{f(n)+A} + B, \quad \text{for} \quad n=1,2,\ldots \end{equation} where $A$ and $B$ are two positive constants. Is it possible to solve these exactly and/or approximately (eg, for large $n$) and obtain a form for $f(n)$, assuming $f(1)$ is known?
Any input would be appreciated as I know very little about recursive equations.
You can write it in the following way:
$$u_{n+1}=\frac{u_n+A}{Bu_n+C} \ \text{with} \ C:=AB+1$$
This kind of recurrence is called a homographic sequence, and there are standard ways to express its general term ; see for example here explaining the connection with geometric or arithmetic sequences. In your case, the "fixed point equation" is given by the quadratic:
$$Bx^2+ABx-A=0$$
has always 2 real roots because $\Delta=(AB)^2+4AB >0$.
A matrix treatment is also possible: see this old answer of mine here.
$$\frac 1 {f_{n+1}}=\frac 1 {f_n+a}+b\implies f_{n+1}=\frac{f_n+a } {b\,f_n+(1+ab) }=\frac{\frac 1bf_n+\frac ab } {f_n+\frac{1+ab}b }$$
Let $m=\frac 1b$, $x=\frac ab$, $y=\frac{1+ab}b$ to make $$f_{n+1}=\frac {m\,f_n+x}{f_n+y }$$ and have a look to my answer to this question where all steps are detailed.
