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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 $u_n\le(1-a)^n\forall n\in\mathbb{N}$?

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Let us show that for every $n\ge 0$, (i) $u_n>0$, (ii) $u_{n+1}<u_n<2u_{n+1}$ and (iii) $u_{n+1}\le (1-a)^{n+1}$ by induction. Clearly all the statements hold for $n=0$. Suppose that all the statements hold for some $n\ge 0$. Then (i) $u_{n+1}>\frac{u_n}{2}>0$, (ii) $$\frac{u_{n+1}}{2}<(1-2a)u_{n+1}<u_{n+1}-au_n<u_{n+1}\Longrightarrow u_{n+2}<u_{n+1}<2u_{n+2},$$ and (iii) $$u_{n+2}=u_{n+1}-au_n<(1-a)u_{n+1}\le (1-a)^{n+2}.$$ The induction is completed.

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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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