recurrence relation limit $x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$

Question

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.

Answer 1

As you point out this is a linear 2nd order homogeneous difference equation with constant coefficients. the general solution is obtained from the roots of the characteristic polynomial, which are $\lambda_1 = a$ and $\lambda_2 = b$ (don't know how did you come up with your $x_1, x_2$...). This way the general solution becames simply $x_n = c_1 a^n + c_2 b^n$. Finally, using the fact that we have $x_0=1$ and $x_1=a+b$, we conclude that $c_1=\frac{-a}{b-a}$ and $c_2=\frac{b}{b-a}$, and so

$$ x_n = -\frac{a^{n+1}}{b-a} + \frac{b^{n+1}}{b-a} . $$

Answer 2

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?

Answer 3

Hint

No need to a determinant! As you've already concluded, the general term of the sequence would become$$x_n={a^{n+1}-b^{n+1}\over a-b}$$Now, the limits can be easily calculated...