How to solve this limit $\lim\limits_{n\rightarrow\infty}\,{{\left( \frac{{{u}_{n}}}{{{v}_{n}}} \right)}^{2040}}$? Let $(u_n)_n$ and $(v_n)_n$ be positive numerical sequences such that $(1+\sqrt{2})^n=u_n+v_n\sqrt{2}$  with $n\in\mathbb{N}$ 
How to determine the limit of 
$$\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \frac{{{u}_{n}}}{{{v}_{n}}} \right)}^{2040}}$$
I start think to find such recessive sequence for $u_n$ terms in order to separate them using induction but this leads to nothing . so as result is there any shortcut to attack this problem?

$(1,1)$ is a solution of the following Pell like equation (or rather negative Pell equation).
$$x^2-2y^2=-1$$ 
but then
$$\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)=-1 \Rightarrow
\left(1+\sqrt{2}\right)^n\left(1-\sqrt{2}\right)^n=(-1)^n \Rightarrow \\
\left(u_n+v_n\sqrt{2}\right)\left(u_n-v_n\sqrt{2}\right)=(-1)^n \Rightarrow
u_n^2-2v_n^2=(-1)^n$$
$(u_n,v_n)$ are integers (easy to show using binomial theorem). So
$$\left(\frac{u_n}{v_n}\right)^2=2+\frac{(-1)^n}{v_n^2} \iff 
\left(\frac{u_n}{v_n}\right)^{2040}=\left(2+\frac{(-1)^n}{v_n^2}\right)^{1020}$$
We also have
$$v_n =\frac{\left(1+\sqrt{2}\right)^n-\left(1-\sqrt{2}\right)^n}{2\sqrt{2}}\Rightarrow
v_n^2=\frac{\left(3+2\sqrt{2}\right)^{n}+\left(3-2\sqrt{2}\right)^{n}-2\cdot(-1)^n}{8}$$
and


*

*$0<3-2\sqrt{2}<1 \Rightarrow \lim\limits_{n\rightarrow\infty}\left(3-2\sqrt{2}\right)^{n}=0$  

*$3+2\sqrt{2}>1 \Rightarrow \lim\limits_{n\rightarrow\infty}\left(3+2\sqrt{2}\right)^{n}=+\infty$
as a result (and because $f(x)=x^{1020}$ is continuous)
$$\lim\limits_{n\rightarrow\infty}\left(\frac{u_n}{v_n}\right)^{2040}=
\lim\limits_{n\rightarrow\infty}\left(2+\frac{(-1)^n}{v_n^2}\right)^{1020}=\left(2+\lim\limits_{n\rightarrow\infty}\frac{(-1)^n}{v_n^2}\right)^{1020}=2^{1020}$$
