Limit of a sequence of fractionsI have that: S n = b ⋅ S n...

Assume that $S_n=\frac{p_n}{q_n}$ is associated with $(p_n,q_n)\in {\mathbb{R}}^2$. The recurrence
$\begin{array}{}\text{(1)}& S_n=\frac{(a+b)S_{n-1}+ab}{S_{n-1}+b}\end{array}$
can be written in the following form
$\begin{array}{}\text{(2)}& \left(\begin{array}{c}{p}_n\\ q_n\end{array}\right)=\left(\begin{array}{cc}a+b& ab\\ 1& b\end{array}\right)\left(\begin{array}{c}{p}_{n-1}\\ q_{n-1}\end{array}\right)\end{array}$
hence the closed form of both the $\{p_n{\}}_{n\ge 0}$ and the $\{q_n{\}}_{n\ge 0}$ sequences just depends on the diagonalization of $M=\left(\begin{array}{cc}a+b& ab\\ 1& b\end{array}\right)$, whose eigenvalues are given by ${\zeta }_{\pm }=\frac{(a+2b)\pm \sqrt{a^2+4ab}}{2}$.
In particular
$\begin{array}{}\text{(3)}& S_n=\frac{A{\zeta }_{+}^n+B{\zeta }_{-}^n}{C{\zeta }_{+}^n+D{\zeta }_{-}^n}\end{array}$
for some constants $(A,B,C,D)$ depending on the initial values. By taking the limit as $n\to +\mathrm{\infty }$, we simply get $\frac{A}{C}$. On the other hand, by assuming $S_n\to L$ as $n\to \mathrm{\infty }$, such limit has to fulfill
$\begin{array}{}\text{(4)}& L=a+\frac{bL}{b+L}\end{array}$
hence the only chances are $L=\frac{1}{2}(a\pm \sqrt{a^2+4ab})$