<s>[INST] Is $u_n\le(1-a)^n\forall n\in\mathbb{N}$? Consider the sequence $\{u_n\}$ where $u_0=1,u_1=1-a$ for some $0< a < 1/4$, and $u_{n+2} = u_{n+1}-au_n$. Is $u_n\le(1-a)^n\forall n\in\mathbb{N}$?
 [/INST] Let $u_n = (1-a)^n b_n$
Then we have that $b_0 = 1$ and $b_1 = 1$.
Now $$b_{n+2} = \frac{b_{n+1}}{1-a} - \frac{ab_n}{(1-a)^2}$$
And so
$$b_{n+2} - b_{n+1} = \frac{ab_{n+1}}{1-a} - \frac{ab_n}{(1-a)^2} = \frac{a}{1-a}\left(b_{n+1} - \frac{b_n}{1-a}\right)$$
Now if $b_{n+1} \le b_n \le \frac{b_n}{1-a}$, we must have that $b_{n+2} \le b_{n+1}$
And so $b_n$ is a decreasing sequence and thus $b_n \le 1$.
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