How would you show sequence terms are bound?

Question

If $a_1 = 4$ and $a_{n+1} = 3 \sqrt{a_n - 2}$ for $n \ge 1$, how would you prove $3 < a_n < a_{n+1} < 6$? I was thinking of using induction, but I don't see how induction can work with relations and multiple of them. Should I split each of them up? As in first prove $3 < a_n$ and work from there?

Answer 1

HINT: Induction works fine. If $3<a_n<6$, then $1<a_n-2<4$, so $1<\sqrt{a_n-2}<2$.

To show that the sequence is increasing, note that

$$\frac{a_{n+1}}{a_n}=3\sqrt{\frac1{a_n}-\frac2{a_n^2}}\;.$$

Find the best possible lower bound for $\dfrac1{a_n}-\dfrac2{a_n^2}$, remembering that we know that $3<a_n<6$.

Answer 2

You can get a base case pretty easily. Now assume $3<a_{n}<a_{n+1}<6$. Note that $$3<a_{n+1} < 6 \implies 1<a_{n+1}-2 < 4 \implies 1<\sqrt{a_{n+1}-2} < 2 \implies 3<3\sqrt{a_{n+1}-2} < 6 \implies 3<a_{n+2} < 6$$ You can use a similar argument to show that $a_{n+1} < a_{n+2}$