Examine the convergence of a sequence $\{a_{n}\}$ which is given by $a_{1}=a>0,a_{2}=b>0, a_{n+2}=\sqrt{a_{n+1}a_{n}},n\ge 1$ I used inequality between arithmetic and geometric means to show that a sequence $\{a_{n}\}$ is bounded:
$$a_{n+2}=\sqrt{a_{n+1}a_{n}}\le \frac{a_{n+1}+a_{n}}{2}$$
Solving this, I get quadratic inequality
$$a^2_{n+1}-2a_{n+1}a_{n}+a^2_{n}\ge 0$$
which gets me to $$0\le a_{n}\le 1$$
thus, sequence is bounded.
I get that sequence is not monotonic, because
$$a_{n}\le a_{n+2} \le {a_{n+1}}$$
or
$$a_{n+1}\le a_{n+2} \le {a_{n}}$$
Is this right?
 A: Let $c_n=\log a_n$. Then the sequence $\{c_n\}_{n\in\mathbb N}$ satisfies the recursive relation:
$$
c_{n+2}=\frac{1}{2}c_{n+1}+\frac{1}{2}c_{n}, \quad c_1=\log a,\,\,c_2=\log b.
$$
Then we have
$$
c_{n+2}-c_{n+1}=-\frac{1}{2}\big(c_{n+1}-c_{n})=\cdots=\frac{(-1)^n}{2^n}(c_2-c_1). \tag{1}
$$
Also
$$
c_{n+2}+\frac{1}{2}c_{n+1}=c_{n+1}+\frac{1}{2}c_{n}=\cdots=c_{2}+\frac{1}{2}c_{1}.
\tag{2}
$$
Combining $(1)$ and $(2)$ we obtain
$$
c_{n+2}=\frac{1}{3}\left(2c_2+c_1+\frac{(-1)^n}{2^n}(c_2-c_1)\right)
$$
and hence
$$
a_n=a^{1/3}b^{2/3}\left(\frac{b}{a}\right)^{(-1)^n/2^{n-2}}\to a^{1/3}b^{2/3},
$$
since $c^{1/2^n}\to 1$, for every $c>0$.
Note. Proving convergence, without finding the limit, is simpler, as $\{a_{2n+1}\}$ is increasing, $\{a_{2n}\}$ is decreasing, and $a_{2n+2}/a_{2n+1}=\sqrt{a_{2n}/a_{2n-1}}=(a_2/a_1)^{1/2^n}\to 1$.
A: *

*Assume WLOG $a<b$.

*Now show that the even subsequence $\{a_{2n}\}$ and the odd subsequence $\{a_{2n-1}\}$ are individually monotonic.

*Show that all the terms of one of these subsequences are greater than all the terms of the other subsequence.

*So you have got bounded monotonic even subsequence and bounded monotonic odd subsequence. One of them is increasing and the other is decreasing. Show that the limits of these two subsequences are equal.

*Conclude that this information is sufficient to show that the whole sequence converges to the same limit.

A: A shortcut: assume that a real sequence is given by some values of $c_1,c_2$ and the recurrence relation $c_{n+2}=\frac{c_n+c_{n+1}}{2}$. Then it is convergent since:
$$ \left|c_{n+1}-c_n\right|=\frac{1}{2}\left|c_{n}-c_{n-1}\right| \tag{1}$$
and it converges to $\frac{c_1+2c_2}{3}$ since:
$$ c_{n}+2 c_{n+1} = c_{n+1}+2 c_{n+2}.\tag{2} $$
By setting $c_n=\log a_n$, it follows that:
$$ \lim_{n\to +\infty} a_n = \sqrt[3]{a_1 a_2^2}.\tag{3}$$
