# If $a_1=\tan\alpha$ and $a_{n+1}=a_n^2\cos^2\alpha+\sin^2\alpha$, show $(a_n)$ is monotone and bounded; compute related limits

Define $$(a_n)_{n \ge 1}$$, such that $$a_1=\tan \alpha$$, $$\alpha \in \left(\frac{\pi} {4},\frac{\pi}{2} \right)$$ (where $$\alpha$$ is fixed), and $$a_{n+1}=a_n^2 \cdot \cos^2 \alpha +\sin^2 \alpha$$ $$\forall n \in \mathbb{N}$$.

a. Prove that $$(a_n) _{n \ge 1}$$ is monotonous and bounded.
b. Compute $$\lim_{n\to \infty} a_n$$ and $$\lim_{n\to \infty}n(a_n-1)$$.

I tried to divide the recurrence relation by $$a_n$$ to prove monotony, but it didn't work. I can't really guess what are $$a_n$$'s bounds either.

By mathematical induction, one can verify that $$1\le a_n \le \tan^2\alpha$$ for all $$n.$$ In addition, one can calculate $$a_n-a_{n+1} = -a_n^2\cos^2\alpha+a_n-\sin^2\alpha = -\cos^2\alpha(a_n-1)(a_n-\tan^2\alpha) \ge 0.$$ Therefore, the sequence $$\{a_n\}$$ is decreasing and bounded and it converges to 1. Now, Observe that $$a_{n+1} - 1 = \cos^2\alpha(a_n^2-1) = (a_n-1)(\cos^2\alpha)(a_n+1) \le k(a_n-1),$$ where $$0 < k=\cos^2\alpha(a_N+1) < 1, n>N.$$ Note that we can find such (sufficiently large) positive integer $$N$$ since $$a_n\to 1$$ as $$n\to\infty$$ and $$2\cos^2\alpha <1.$$ Thus, the sequence $$a_n-1 \le k^{n-N}(a_N-1)$$ and $$n(a_n-1)\to 0$$ as $$n\to\infty.$$