TITLE: If $(1+\sqrt{2})^n=a_{n}+b_{n}\sqrt{2}\;,(\forall n\in \mathbb{N}).$Then $\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = $
QUESTION [6 upvotes]: Let $a_{n},b_{n}\in \mathbb{Q}$ such that $(1+\sqrt{2})^n=a_{n}+b_{n}\sqrt{2}\;,(\forall n\in \mathbb{N}).$Then $\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = $

$\bf{My\; Try::}$ Using 
$$(1+\sqrt{2})^n = 1+\binom{n}{1}(\sqrt{2})+\binom{n}{2}(\sqrt{2})^2+\cdots \cdots +\binom{n}{n}(\sqrt{2})^n$$
So $$a_{n} = \binom{n}{0}+\binom{n}{2}\cdot 2+\binom{n}{4}\cdot 2^2+\cdots \cdots $$
So $$b_{n} = \binom{n}{1}+\binom{n}{3}\cdot 2+\binom{n}{5}\cdot (2)^2+\cdots$$
Now How can i solve $$\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = $$
Help required, Thanks

REPLY [2 votes]: Note first that the recursion
$$a_{n+1}+b_{n+1}\sqrt2=(1+\sqrt2)(a_n+b_n\sqrt2)=(a_n+2b_n)+(a_n+b_n)\sqrt2$$
implies $a_1,a_2,a_3,\ldots$ and $b_1,b_2,b_3,\ldots$ are both strictly increasing sequences of positive integers, since $a_1=b_1=1$ and $a_n,b_n\gt0$ implies $a_{n+1}=a_n+2b_n\gt a_n$ and $b_{n+1}=a_n+b_n\gt b_n$.  In particular, $b_n\to\infty$ as $n\to\infty$.
Now, by the formal symmetry between $\sqrt2$ and $-\sqrt2$ and the general algebraic identity $(x+y)(x-y)=x^2-y^2$, we have
$$\begin{align}
(1+\sqrt2)^n=a_n+b_n\sqrt2&\implies(1-\sqrt2)^n=a_n-b_n\sqrt2\\
&\implies(-1)^n=a_n^2-2b_n^2\\
&\implies{a_n\over b_n}=\sqrt{2\left(1+{(-1)^n\over b_n^2}\right)}\\
&\implies\lim_{n\to\infty}{a_n\over b_n}=\sqrt2
\end{align}$$