limit of a recursive sequence:2 Let $$x_k = \frac{A}{1-C} x_{k-1} + \frac{B}{1-C}x_{k-2},$$ where $A, B, C$ are positive reals such that $A + B + C =1$. Let $$x_1 = 1$$ and $$x_2 = 1 + y,$$ with $y$ is positive. Which conditions should I impose on the parameters $A$, $B$, $C$, $y$ such that the sequence $x_k$ goes to infinite as $k$ goes to infinite (if this is possible) ?

Let $a=\frac{A}{1-C}$ and $b=\frac{B}{1-C}$; then $a+b=1$. As was noted in the comments, this makes $x_k$ a convex combination of $x_{k-1}$ and $x_{k-2}$, and since $a,b>0$, $x_k$ lies strictly between $x_{k-1}$ and $x_{k-2}$. Thus, $1\le x_k\le 1+y$ for all $k\ge 1$, and the sequence is bounded.
The characteristic polynomial of the recurrence $x_k=ax_{k-1}+bx_{k-2}$ is $$t^2-at-b=t^2-at-(1-a)\;,$$ whose roots are $1$ and $a-1$. Thus, $x_k=\alpha(1^k)+\beta(a-1)^k=\alpha+\beta(a-1)^k=\alpha+\beta(-b)^k$ for some $\alpha$ and $\beta$. $1=x_1=\alpha-b\beta$, and $1+y=\alpha+b^2\beta$, so $(b^2+b)\beta=y$, $$\beta=\frac{y}{b^2+b}\;,$$ and $$\alpha=1+b\beta=1+\frac{y}{b+1}\;.$$ Thus,
$$x_k=1+\frac{y}{b+1}+\frac{(-1)^kb^ky}{b(b+1)}=1+\frac{y}{b+1}\left(1+(-1)^kb^{k-1}\right)$$
for all $k\ge 1$. Clearly $$\lim_{k\to\infty}x_k=1+\frac{y}{b+1}\;,$$ since $|b|<1$.
