I have a good understanding of the theorems of convergent and divergent sequence and I am able to apply them. But I really don't see how I can solve this thing. We have to use the theorem provided as well in order to solve the question.

Problem

Consider a real sequence $$(a_n)$$ defined through the formula $$c_2 a_{n+2} + c_1a_{n+1} + x_0a_n = 0 \tag{1}$$ with $$c_2, c_1, c_0 \in \Bbb{R}$$. The characteristic polynomial of the linear recurrence $$(1)$$ is defined to be $$p(x) = c_2 x^2 + c_1 x + c_0.$$ By means of the following theorem (do not prove it!)

Theorem If $$p(x) = c_2(x - x_1)(x - x_2)$$ where $$x_1$$ and $$x_2$$ are distinct nonzero complex numbers, then $$(1)$$ can be solved for $$a_n$$ and we have $$a_n = B_1x_1^n + B_2x_2^n \quad \forall n \in \Bbb{N} \quad \text{with} \quad B_1, B_2 \in \Bbb{C} \tag{2}$$

1. find $$B_1$$ and $$B_2$$ for the sequence $$(a_n)_{n\in\Bbb{N}}$$ such that $$a_0 = a, \quad a_1 = b, \quad a_n = \frac{a_{n-1} + a_{n-2}}{2} \quad \forall n \ge 2$$ with $$a, b \in \Bbb{R}$$.

2. Moreover, show that the above sequence converges and find its limit.

I tried to re organize the equation of $$a_n=\frac{a_{n-1} +a_{n-2}}{2}$$ to $$a_n-\frac{1}{2}a_{n-1}-\frac{1}{2}a_{n-2}=0$$ but then I didn't know how to take the choice for $$x$$ or how to proceed after that.

With your reworked equation we have that the characteristic polynomial is given by $$p(x)=x^{2}-\frac{1}{2}x-\frac{1}{2}=(x-1)(x+\frac{1}{2}).$$ By the theorem this implies there exist $$B_{1},B_{2}\in\mathbb{C}$$ such that $$a_{n}=B_{1}+B_{2}(-2)^{-n}.$$ In particular we have $$B_{1}+B_{2}=a$$ and $$B_{1}-\frac{1}{2}B_{2}=b$$. This gives us $$B_{1}=\frac{1}{3}(a+2b)$$ and $$B_{2}=\frac{2}{3}(a-b)$$.
Furthermore note that $$\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}B_{1}+B_{2}(-2)^{-n}=B_{1}=\frac{1}{3}(a+2b).$$
• $x_{2}=-\frac{1}{2}=(-2)^{-1}$. Oct 16 '19 at 15:22