For which values of $\alpha,\beta$ does this series converge?

Question

I have found the general solution for a reccurence relation:

$$z_n=\alpha(1+\sqrt{3})^n+\beta(1-\sqrt{3})^n,$$

and is asked for which values of $\alpha$ and $\beta$ it converges. My first thought is that it converges for $\alpha=0$ and $\beta\in\mathbb{R}$, as $(1+\sqrt{3})^n\to\infty$ and $(1-\sqrt{3})^n\to0$, however, it seems too simple. Any thoughts on this would be greatly appreciated.

Answer 1

Suppose that $(z_n)$ is convergent, then the sequence $(z_n-\beta(1-\sqrt{3})^n)$ is convergent for each $ \beta$. But then $(\alpha(1+\sqrt{3})^n)$ is convergent. This is the case $ \iff \alpha =0$.

Result: $(z_n)$ converges $ \iff \alpha =0$ and $\beta$ is arbitrary.

Answer 2

We have that for $\alpha \neq 0$

• $1+\sqrt{3}>1 \implies |\alpha(1+\sqrt{3})^n| \to \infty $

and for any $\beta$

• $|1-\sqrt{3}|<1\implies \beta(1-\sqrt{3})^n\to 0$

therefore

$$z_n=\alpha(1+\sqrt{3})^n+\beta(1-\sqrt{3})^n\to 0 \iff \alpha =0$$

otherwise the sequence diverges.

When $\alpha=0$ also the associated geometric series converges to

$$\sum_{1}^\infty z_n=\frac{\beta}{1-(1-\sqrt 3)}=\frac{\beta}{\sqrt 3}$$

Answer 3

Apparently, it is the recurrence relation: $z_{n+2}=2z_{n+1}+2z_n$, whose characteristic equation is: $x^2-2x-2=0$ and characteristic roots are: $x_1=1+\sqrt{3}$ and $x_2=1-\sqrt{3}$.

The solution $z_n=\alpha (1+\sqrt{3})^n+\beta(1-\sqrt{3})^n$ is convergent to zero if and only if its characteristic roots are satisfied: $|x_1|<1$ and $|x_2|<1$.

Since $|x_1|=1+\sqrt{3}>1$, then it implies $\alpha =0$. Since $|x_2|=|1-\sqrt{3}|<1$, then it implies $\beta \in \mathbb R$. See the reference.