Limits of Recurrence Relations with $0<b<a$

[Vali]

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 $L= ?$
b) If $0<b<a<1$ and $L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}$ then $L= ?$
I don't know how to start.I tried to write the first terms $x_{2},x_{3}...$ but I didn't get too far.

 

[MountEvariste]

Define $\displaystyle f(y) = \sum_{n \ge 0} x_n y^n$. Multiply both sides of your relation by $y^n$ and sum over $n \ge 0$:

$\displaystyle \mathcal{LHS} = \sum_{n \ge 0}x_{n+2}y^n = \frac{1}{y^2}\left(\sum_{ n \ge 0} x_n y^n - x_0y^0-x_1y^1\right) = \frac{1}{y^2}\left(f(y) - (a+b)y-1\right)$

$\displaystyle \mathcal{RHS} = (a+b)\sum_{n \ge 0}x_{n+1}y^n-(ab)\sum_{n \ge 0}x_{n}y^n = \frac{a+b}{y} \left(f(y)-1\right)-ab f(y), $ and:

$\displaystyle \mathcal{LHS} = \mathcal{RHS} \implies f(y) = \frac{1}{(ay - 1)(by-1)} = \sum_{j \ge 0}a^j y^j \cdot \sum_{\ell \ge 0}b^{\ell}y^{\ell} = \sum_{n \ge 0}\sum_{0 \le j \le n}a^j y^j b^{n-j}y^{n-j}$

Since the inner sum equals $(a-b)^{-1}(a^{n+1}-b^{n+1})y^n$ going back to the definition of $y(n)$ we've:

$\displaystyle \sum_{n \ge 0}{x_n y^n} = \sum_{n \ge 0} \frac{a^{n+1}-b^{n+1}}{a-b} y^n $ and we can readily read off from the coefficient: $\displaystyle x_n = \frac{a^{n+1}-b^{n+1}}{a-b}.$

Can you do (a) and (b) now? You can also find $x_n$ by solving the characteristic equation.

 

[Vali]

I solved both limits!I used the characteristic equation to find $$x_{n}$$
Thanks!