# Limit of a sequence defined by recursive relation : $a_n = \sqrt{a_{n-1}a_{n-2}}$

We're given a sequence defined by the recursive relation: $$a_n = \sqrt{a_{n-1}a_{n-2}}$$

$a_1$ and $a_2$ are positive constants. We have to show the following:

1. The sequences $\{ b_n \} = \{ a_{2n-1} \}$ and $\{ c_n \} = \{ a_{2n} \}$ are monotonic, and if one is increasing, the other is decreasing.

2. The limit of the sequence $\{ a_n \}$ is $\left(a_1a_2^2\right)^{\frac13}$

Now, I have proved the first part. Besides that, I have also proved a few other things:

If $a_1 > a_2$, then:

a. $\{ b_n \}$ decreases, and $\{ c_n \}$ increases.

b. $c_n < b_n$

If $a_1 < a_2$, we just flip $\{ b_n \}$ and $\{c_n\}$

Besides, I have also shown that both the sequences : $\{b_n\}$ and $\{ c_n\}$ have the same limit. What I don't know, is how to evaluate the limit.

• What is $a_{n+1}^2\cdot a_n$? – Daniel Fischer Oct 15 '14 at 17:42
• Wow. Okay, I think this will be enough. Thanks! – Parth Thakkar Oct 15 '14 at 17:43

Using the relation $a_{n} = \sqrt{a_{n-1}a_{n-2}}$ for $n \geqslant 2$, we find that
$$a_{n+1}^2a_n = (\sqrt{a_na_{n-1}})^2a_n = a_na_{n-1}a_n = a_n^2a_{n-1}$$
is independent of $n$, so $a_{n+1}^2a_n = a_2^2a_1$ for all $n$, and hence $\lambda^3 = a_2^2a_1$ for $\lambda = \lim\limits_{n\to\infty} a_n$.