Show that the sequence is convergent to $(a_1a_2^2)^{1/3}$. 
If $\{ a_n\}$ be a sequence of positive real numbers such that $$a_n=\sqrt{a_{n-1}a_{n-2}},n>2$$ then show that the sequence is convergent to $(a_1a_2^2)^{1/3}$. 

How I proceed: Take square of $a_n$ so I get $a_n^2=a_{n-1}a_{n-2} \implies a_n^3=a_na_{n-1}a_{n-2}$. Then stuck. Please help.
 A: Let $b_n=\ln (a_n)$
So, $2b_n=b_{n-1}+b_{n-2}$
Using this, $b_n=A+B\left(-\frac12\right)^n$ where $A,B$ are arbitrary constants
$n=1\implies b_1=A-\frac B2$
$n=2\implies b_2= A+\frac B4$
$\implies B=\frac43(b_2-b_1)$ and $A=\frac{b_1+2b_2}3$
$\implies b_n=\frac{b_1+2b_2}3+\frac43(b_2-b_1)\left(-\frac12\right)^n$
$\implies\lim_{n\to\infty} b_n=\frac{b_1+2b_2}3$ as $\lim_{n\to\infty}r^n=0$ if $|r|<1$
$\implies\lim_{n\to\infty} \ln(a_n)=\frac{\ln a_1+2\ln a_2}3=\ln (a_1\cdot a_2^2)^{\frac13} $
A: Let $r$ be a positive real number, and let $b_n=ra_n$.  If one sequence has a limit, then so does the other, with the limits differing by a factor of $r$.  Note that the sequence of $b$'s satisfies the same recursion as the $a$'s:  
$$b_n=ra_n=r\sqrt{a_{n-1}a_{n-2}}=\sqrt{(ra_{n-1})(ra_{n-2})}=\sqrt{b_{n-1}b_{n-2}}$$
Now let $r=(a_1a_2^2)^{-1/3}$.  Then we have
$$b_1={a_1\over(a_1a_2^2)^{1/3}}=\left({a_1\over a_2} \right)^{2/3} $$
and
$$b_2={a_2\over(a_1a_2^2)^{1/3}}=\left({a_2\over a_1} \right)^{1/3}= b_1^{-1/2}$$
Let's just write $b$ for $b_1$.  Then by the recursion, we have
$$\begin{align}
b_3&=\sqrt{b_2b_1}=(b^{-1/2}b)^{1/2}=b^{1/4}\\
b_4&= \sqrt{b_3b_2}=(b^{1/4}b^{-1/2})^{1/2}=b^{-1/8}\\
b_5&=\sqrt{b_4b_3}=(b^{-1/8}b^{1/4})^{1/2}=b^{1/16}\\
&\vdots
\end{align}$$
It's an easy induction to see that $b_{n+1}=b^{(-1/2)^n}$, and consequently easy to see that $\lim_{n\rightarrow\infty}b_n=1$.  From this it follows that the sequence of $a$'s converges, and
$$\lim_{n\rightarrow\infty}a_n={1\over r} = (a_1a_2^2)^{1/3}$$
