Suppose that a bounded sequence $\{a_n \}_{n \geq 1}$ is such that $$a_{n + 2} \leq \dfrac {1} {3} a_{n+1} + \dfrac {2} {3} a_n,\ \ \ \ \text {for}\ n \geq 1$$ Prove that the sequence $\{a_n \}_{n \geq 1}$ is convergent.
What I find is that for all $n \geq 1,$ we have $$a_{n + 2} – a_2 \leq \dfrac {2} {3} (a_1 – a_{n+1}).$$
Also it can't be eventually monotone increasing since for otherwise for all $n \geq 1,$ we have $$a_{n+2} – a_{n+1} \leq \dfrac {2} {3} (a_n – a_{n+1}) \leq 0 \implies a_{n+2} \leq a_{n+1},$$ a contradiction. So if the sequence is eventually monotone it has to be eventually monotone decreasing.
Is it of any importance? Thanks.
Best Answer
With $c>0$ determined below, let $b_n=a_{n+1}+ca_n$. Then $\{b_n\}_{n\ge1}$ is also bounded from below and we obtain $$b_{n+1}=a_{n+2}+ca_{n+1}\le \left(\frac13+c\right)a_{n+1}+\frac23 a_n=\frac{(1+3c)a_{n+1}+2a_n}{3}.$$ If we smartly pick $c=\frac23$, this amounts to $$ b_{n+1}\le b_n.$$ We conclude that sequence $b_n$ converges to some limit $b$.
By applying $\liminf$ to both sides of the equation $a_{n+1}=b_n-\frac23 a_n$, we find $$\tag1\liminf a_n=b-\frac23\limsup a_n $$ and similarly $$\tag2\limsup a_n=b-\frac23\liminf a_n.$$ Solving the linear equations $(1)$ and $(2)$, we arrive at $$ \liminf a_n=\limsup a_n=\frac 53 b,$$ i.e., $$\lim a_n=\frac 53 b. $$