Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question

2 Answers 2

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.

share|cite|improve this answer

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 $b_n \le 1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.