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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.

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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 $n\to\infty.$

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