<s>[INST] Prove this $|a_{n}|\le 2$ Let sequence such $$|a_{n}-\dfrac{a_{n+1}}{2}|\le 1$$
if $|a_{n}|\le \dfrac{3^n}{2^n}$, show that
$$|a_{n}|\le 2$$
since 
$$2a_{n}-2\le a_{n+1}\le 2a_{n}+2$$
so we 
$$a_{n+1}-2\ge 2(a_{n}-2)\ge\cdots \ge 2^n(a_{1}-2)$$
 [/INST] Suppose $a_N>2$. Put $a_N=2+k$ where $k>0$. Then we have $1+k\le\frac{1}{2} a_{N+1}$ and so $a_{N+1}\ge2+2k$. By a trivial induction $a_{N+m}\ge2+2^mk$. Now we claim that for sufficiently large $m$ we have $a_{N+m}>\frac{3^{N+m}}{2^{N+m}}$.
For $\left(\frac{4}{3}\right)^m>\frac{1}{k}\left(\frac{3}{2}\right)^N$ for sufficiently large $m$ and hence $2^mk>\left(\frac{3}{2}\right)^{N+m}$, which gives the required result since $a_{N+m}>2^mk$. But that gives us a contradiction, since we are given that $a_n\le\left(\frac{3}{2}\right)^n$ for all $n$. So we cannot have $a_N>2$.
Similarly, if $a_N<-2$. for then we can take $a_N=-2-k$ where $k>0$, and in a similar way we get $a_{N+m}<-2-2^mk$ and hence $a_{N+m}<-\left(\frac{3}{2}\right)^{N+m}$ for sufficiently large $m$, which again is a contradiction. So we cannot have $a_N<-2$.
So we have established that we must have $|a_N|\le2$ for all $N$.
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