TITLE: Tips on proving this convergence.
QUESTION [5 upvotes]: We have an inductively defined sequence $x_n=x_{n-1}+2y_{n-1}$ and $y_n=x_{n-1}+y_{n-1}$ where $x_n^2-2y_n^2=\pm 1$, where $x_0=1$ and $y_0=0$.
I need to prove that the sequence $\left(\frac{x_n}{y_n}\right)_{n=1}^\infty$ converges to $\sqrt2$.
Now I can see that $\left(\frac{x_n^2}{y_n^2}\right)_{n=1}^\infty$ converges to $2$, so that may be easier to prove.
It follows now that in order to prove this, $\forall\epsilon >0, \exists n>N$ s.t. $|\frac{x_n^2}{y_n^2}-2|<\epsilon$.
Can anybody point me in the right direction for solving this?

REPLY [2 votes]: The recursion can be written as
$$
\begin{bmatrix}x_n\\y_n\end{bmatrix}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}x_{n-1}\\y_{n-1}\end{bmatrix}
$$
The matrix has two eigenvalues: the roots of $\lambda^2-2\lambda-1=0$; that is, $\lambda=1\pm\sqrt{2}$.
The eigenvector with the eigenvalue of $1-\sqrt{2}$ is
$$
\begin{bmatrix}1&2\\1&1\end{bmatrix}\color{#C00000}{\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}=\begin{bmatrix}-2+\sqrt{2}\\-1+\sqrt{2}\end{bmatrix}=(1-\sqrt{2})\color{#C00000}{\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}
$$
The eigenvector with the eigenvalue of $1+\sqrt{2}$ is
$$
\begin{bmatrix}1&2\\1&1\end{bmatrix}\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}}=\begin{bmatrix}2+\sqrt{2}\\1+\sqrt{2}\end{bmatrix}=(1+\sqrt{2})\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}}
$$
Therefore,
$$
\begin{align}
\begin{bmatrix}x_n\\y_n\end{bmatrix}
&=\begin{bmatrix}x_0\\y_0\end{bmatrix}\begin{bmatrix}1&2\\1&1\end{bmatrix}^n\\
&=\left(a\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+b\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}\right)\begin{bmatrix}1&2\\1&1\end{bmatrix}^n\\
&=a(1+\sqrt2)^n\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+b(1-\sqrt2)^n\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}\\
&=a(1+\sqrt2)^n\left(\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+\frac{b}{a}(2\sqrt{2}-3)^n\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}\right)
\end{align}
$$
Assuming $a\ne0$, what is the limit of $x_n/y_n$ in red?