Evaluate $\lim_{n\to\infty} \frac{b_n}{a_n}$

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, $\lim_{n\to \infty}r_n=\tan (\lim_{n\to \infty}\theta_n)=\tan(\pi/2+\pi/2-\pi/4)=-1.$