<s>[INST] recurrence relation limit $x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$ Let $0<b<a$ and $(x_{n})_{n\in \mathbb{N}}$ with $x_{0}=1, \ x_{1}=a+b$
$$x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$$
a) If $0<b<a$ and $L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}$ then $L= ?$ ($L=a$ right answer )
b) If $0<b<a<1$ and $L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}$ then $L= ?$ ($L=\frac{1}{(1-a)(1-b)}$ right answer)
My try:
First, I tried to write  $x_{2},x_{3}...$ but I didn't get too far.I obtained $x_{2}=(a+b)^{2}-ab$ and $x_{3}=(a+b)^{2}-ab-ab(a+b)$.
I tried  to write the characteristic equation like:
$x_{n+2}-(a+b)x_{n+1}-abx_{n}=0$ so the equation is $r^{2}-(a+b)r-ab=0$.The discriminant is positive so $x_{n}=c\cdot r_{1}^{n}+d\cdot r_{2}^{n}$ where $c$ and $d$ are the solutions of the system: $\begin{cases}
c\cdot r_{1}^{1}+d\cdot r_{2}^{1}=x_{1} \\ 
c\cdot r_{1}^{2}+d\cdot r_{2}^{2}=x_{2}
\end{cases}$ but the calculations are too complicated and I get stuck.
 [/INST] Hint: Try rewriting your recurrence as follows:
$$
x_{n+2} = (a+b)x_{n+1} - abx_n \\
x_{n+2} = ax_{n+1}+bx_{n+1} - abx_n \\
x_{n+2} - bx_{n+1} = a(x_{n+1} - bx_n)
$$
So you get:
$$
x_{n+2} - bx_{n+1} = a(x_{n+1} - bx_n) = a^2(x_{n} - bx_{n-1}) = \cdots
$$
Can you proceed from here?
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