Given a recurrence relation: $$ x_{n+2} = {5\over 2}x_{n+1} + x_n\\ x_1 = a\\ n\in\Bbb N $$ Find the value of $x_2$ for $x_n$ to be convergent and evaluate: $$ \lim_{n\to\infty}x_n $$
First assume the limits exists, then it must follow: $$ \lim_{n\to\infty}x_n = L \implies 2L = 5L + 2L \iff L = 0 $$
The only possible value for the limit is $L = 0$.
Then I proceeded with writing down a characteristic equation for the recurrence: $$ \lambda^2 = {5\over 2}\lambda + 1 \\ 2\lambda^2 = 5\lambda + 2\\ 2\lambda^2 - 5\lambda - 2 = 0 $$
So the roots are given by: $$ \lambda = \frac{5 \pm \sqrt{41}}{4} $$
Given that fact the closed form of the recurrence is as follows: $$ x_n = c_1\lambda_1^n + c_2\lambda_2^n $$
Or writing down the initial conditions: $$ c_1\lambda_1^1 + c_2\lambda_2^1 = a \\ c_1\lambda_1^2 + c_2\lambda_2^2 = x_2 $$
From here I'm not sure where to go. $c_1, c_2$ are still unknown and will remain unknown unless $x_2$ is given which is a circular dependency. The answer section suggests that: $$ x_2 = a\frac{5-\sqrt{41}}{4} $$
How do I obtain that result? Please note this problem is from limits section, and I would like to stick to the simplest possible solution. Thank you!
Let $\lambda_1 > 0$ and $\lambda_2 < 0$. It's clear from OP's finding $$\lambda = \frac{5 \pm \sqrt{41}}{4}$$ that $\lambda_1 \in \left(\cfrac{11}{4},3\right)$ and $\lambda_2 \in \left(-\cfrac12,-\cfrac14\right)$. Therefore, $\lambda_1^n \to \infty$ and $\lambda_2^n \to 0$ as $n\to\infty$. Since $(x_n)_n$ has to be convergent, $c_1 = 0$. The system of linear equations $$c_1\lambda_1^1 + c_2\lambda_2^1 = a \\ c_1\lambda_1^2 + c_2\lambda_2^2 = x_2$$ then becomes \begin{cases} c_2\lambda_2 &= a \\ c_2\lambda_2^2 &= x_2. \end{cases}
Then $c_2 = x_2 / \lambda_2^2$ and $a = c_2\lambda_2 = x_2 / \lambda_2$. Hence $$x_2 = \lambda_2 a = -\frac{\sqrt{41}-5}{4} a$$