Let $\alpha$ and $\beta$ be the roots of

Question

Let $\alpha$ and $\beta$ be the roots of $x^2-4x+2=0$ and let $t_n$=$\frac{\alpha^{n+1}+\beta^{n+1}}{\alpha^n+\beta^n}$ for all $n\geqslant1$. Evaluate $\lim: \lim_{n\to \infty}t_n$.

I found roots: $2+\sqrt2$ and $2-\sqrt2$ but now I don’t know what to do

Answer 1

Set $\alpha=2-\sqrt{2}$ and $\beta=2+\sqrt{2}.$

Since $\alpha=2-\sqrt{2}$ is less than $1$, $\alpha^n$ and $\alpha^{n+1}$ will go to $0$ as $n \rightarrow \infty$. $$\lim_{n \rightarrow \infty} \frac{\alpha^{n+1}+\beta^{n+1}}{\alpha^n+\beta^n}=\lim_{n \rightarrow \infty} \frac{\beta^{n+1}}{\beta^n}=\beta.$$

Answer 2

Hint: One of the roots is between $0$ and $1$, while the other root is greater than $1$.

Answer 3

The roots are $\alpha=2-\sqrt{2}$ and $\beta=2+\sqrt{2}$. Therefore $\alpha^n\to 0$ and $\beta^n\to \infty$. Now write

$$t_n=\beta\cdot{\left({\alpha\over\beta}\right)^{n+1}+1\over\left({\alpha\over\beta}\right)^n+1}$$

And we get that $t_n\to\beta$

Answer 4

Let $x_n = \alpha^n+\beta^n$. Then $x_{n+2} = 4x_{n+1} - 2x_n$ and so $$ t_{n+1} = \frac{x_{n+2}}{x_{n+1}} = \frac{4x_{n+1} - 2x_n}{x_{n+1}} = 4 - \frac{2}{t_n} $$ If $t_n \to L$, then $L^2 = 4L-2$ and $L=2\pm\sqrt{2}$.

By induction, $t_n \ge 2$. Therefore, $L \ge 2$ and so $L=2+\sqrt{2}$.