Determine values for which the general solution converges

Question

Textbook problem.

Given the following general solution to a recurrence relation $$z_n = \alpha(1+\sqrt{3})^n + \beta(1-\sqrt{3})^n$$ For which values $\alpha, \beta$ does the solution converge? And determine the order of the rate of convergence for these values.

By attempting to plot the sequence in some interval with varying values of $\alpha, \beta$ it seems like it will converge whenever $\alpha=0$ and $\beta = (-\infty, \infty)$, but how can i go about determining this in a more rigorous way?

Answer 1

You are correct.

For $a \neq 0$ we have that $z_n \to +\infty$ or $-\infty$ since $(1+\sqrt{3})>1$

Also $(1-\sqrt{3})<1 $

Thus the sequence converges $\forall b \in \Bbb{R}$ and for $a=0$

And converges to zero ,for every such value.

Answer 2

Since $z_n$ coverges it is bounded, so the sequence $${z_n\over (1+\sqrt{3})^n}$$ also coverges and it limit is $0$, so:

$$ 0 = \alpha +\beta \lim_{n\to \infty}{(1-\sqrt{3})^n\over (1+\sqrt{3})^n}= \alpha+\beta\cdot 0\implies \alpha =0$$ and clearly $\beta $ can be any real number.

Answer 3

$$(1+\sqrt 3)^n$$ diverges and $$(1-\sqrt 3)^n$$ converges to zero.