What is the exact limit of the following monotone sequence?

Question

$$1\le a_1\le 2,a_{n+1}^2=3a_n-2,n\in\Bbb R$$ Is the sequence monotone decreasing or increasing?

Answer 1

Hint:  compare $\,a_2\,$ to $\,a_1\,$, then note that $\;a_{n+1}^2-a_n^2=3(a_n-a_{n-1})\,$ so the difference between consecutive terms preserves the same sign.

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[ EDIT ]  Expanded hint: $\,a_2^2-a_1^2=3a_1-2-a_1^2 = -(a_1-1)(a_1-2) \ge 0\,$.

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[ EDIT #2 ]  The equivalence used above that $\;a^2 \ge b^2 \;\iff\; a \ge b\,$ implicitly assumes that both sides are positive i.e. $\,a_n \ge 0\,$.

Though not explicitly stated in the question, this would be a reasonable assumption to make because (a) the question is tagged as real-analysis and any negative $a_n$ would yield an imaginary $a_{n+1}\,$, and (b) considerations of monotonicity would not apply if complex numbers were allowed.

Answer 2

Hint: First show that $a_{n}\geq0$ for all $n$. Then use $a_{n+1}^{2}-a_{n}^{2}=3(a_{n}-a_{n-1})$

Answer 3

The statement $a_{n+1}^2=3a_n-2$ doe not uniquely define $a_{n+1}.$ I will assume that $a_{n+1}\geq 0.$ This will require showing that $3a_n-2\geq 0$ for every $n,$ in order that the real number $a_{n+1}=\sqrt {3a_n-2}\;$ exists.

(1). $a_n\geq 1\implies 3a_n-2\geq 1\implies a_{n+1}=\sqrt {3a_n-2}\;\geq 1. $ Since $a_1\geq 1,$ we have $a_n\geq 1$ for all $n,$ by induction on $n.$

(2). If $1<a_n<2$ then, since $a_n>0,$ we have: $$a_n<a_{n+1}<2\iff a_n<\sqrt {3a_n-2}\;<2\iff$$ $$\iff a_n^2<3a_n-2<4\iff (a_n^2-3a_n+2<0 \land 3a_n<6)\iff$$ $$\iff ((a_n-1)(a_n-2)<0\land a_n<2).$$ The last line above is equivalent to $1<a_n<2.$

(3). So if $1<a_1<2$ then the sequence is strictly increasing, and $2$ is an upper bound for the sequence, so it has a limit $L ,$ with $2\geq L>1.$ But if $a_n$ converges to $L$ then $a_{n+1}^2$ converges to $L^2,$ so $$L^2=\lim_{n\to \infty}a_{n+1}^2=\lim_{n\to \infty}(3a_n-2)=3L-2.$$ This gives $L^2=3L-2$ with $L>1.$ That is, $L=2.$

(4). If $a_1=2$ then $a_2=2.$ And if $a_1=1$ then $a_2=1.$