Evaluate $\lim_{n\to\infty} \frac{b_n}{a_n}$ Problem:
Two sequence $a_n, b_n$ which satisfy
\begin{cases} a_{n+1}=n^2a_n -2b_n \\ b_{n+1}=n^2b_n +2a_n  \end{cases} and$$a_1 =1, \quad b_1 = 0$$Find $$\lim_{n\to\infty} \frac{b_n}{a_n}$$
How can I approach? I couldn't find any relation of $a_n$ and $b_n$.
 A: Let $r_n=\frac{b_n}{a_n},\ n\ge 1$. Then, we have the iterative sequence, $$r_{n+1}=\frac{r_n+c_n}{1-c_nr_n},\ c_n=2/n^2\\\implies \theta_{n+1}=\theta_n + \arctan \frac{2}{n^2},$$ where $\theta_n= \arctan r_n$. Therefore, $$\theta_n = \theta_1+\sum_{k=1}^n \arctan \frac{2}{n^2}=\sum_{k=1}^n \arctan \frac{2}{n^2}.$$ Now, observe that $\arctan(2/n^2)=\arctan\left(\frac{n+1-(n-1)}{1+(n+1)(n-1)}\right)=\arctan (n+1)-\arctan (n-1)$, so that, $\theta_n = \sum_{k=2}^{n+1}\arctan k-\sum_{k=0}^{n-1}\arctan k=\arctan n + \arctan (n+1)-\frac{\pi}{4}.$ Therefore, $\lim_{n\to \infty}r_n=\tan (\lim_{n\to \infty}\theta_n)=\tan(\pi/2+\pi/2-\pi/4)=-1.$
A: Calling $P_k = \left(\begin{array}{c}a_k\\ b_k\end{array}\right)$ 
$$
V = \frac{1}{\sqrt 2}\left(
\begin{array}{cc}
 1 & i \\
 -1 & i \\
\end{array}
\right),\ \ \ \Lambda_k = \left(
\begin{array}{cc}
 k^2-2i & 0 \\
 0 & k^2+2i \\
\end{array}
\right)
$$
$$
M_k = \bar V^{\dagger}\cdot\Lambda_k\cdot V
$$
we have
$$
\left(V\cdot P_{n+1}\right) = M_k\cdot\left( V\cdot P_n\right)
$$
or calling
$$
R_n = V\cdot P_n
$$
$$
R_{n+1} = M_n\cdot R_n
$$
and thus
$$
R_n = \prod_{k=1}^n M_k\cdot R_0
$$
but
$$
\prod_{k=1}^n M_k = \left(\prod_{k=1}^n \rho_k\right)\left(\begin{array}{cc}e^{i\sum_{k=1}^n \phi_k}& 0 \\ 0 & e^{-i\sum_{k=1}^n\phi_k} \end{array}\right)
$$
then
$$
V\cdot \left(\begin{array}{c}a_k\\ b_k\end{array}\right) = \left(\prod_{k=1}^n \rho_k\right)\left(\begin{array}{cc}e^{i\sum_{k=1}^n \phi_k}& 0 \\ 0 & e^{-i\sum_{k=1}^n\phi_k} \end{array}\right)\cdot V\cdot \left(\begin{array}{c}1\\ 0\end{array}\right)
$$
so
$$
a_n= \frac{1}{2} e^{-i \Phi_n } \left(e^{2 i \Phi }+1\right) \left(\prod_{k=1}^n \rho_k\right) ,b_n= -\frac{1}{2} i e^{-i \Phi_n } \left(e^{2 i \Phi
   }-1\right) \left(\prod_{k=1}^n \rho_k\right)
$$
and
$$
\frac{b_n}{a_n} = \tan{\Phi_n}
$$
with 
$$
\Phi_n = \sum_{k=1}^n\arctan\left(\frac{2}{k^2}\right)
$$
hence
$$
\lim_{n\to\infty}\frac{b_n}{a_n} = -1
$$
