solve the recurrence relation $a_n= \sqrt{a_{n-1} a_{n-2}}$

Question

Solve $a_n= \sqrt{a_{n-1} a_{n-2}}$ for $n \ge 2$ if $a_0=2$ and $a_1=8$.

I tried generating functions like $ f(x)= \sum\limits_{i=0}^{\infty}a_i^2 x^i$ but I couldn't figure it out... any help would be greatly appreciated.

Answer 1

Set $b_n=\ln a_n$. Then, $b_0=\ln 2$, $b_1=\ln 8$ and $$ b_n=\frac{1}{2}(b_{n-1}+b_{n-2}). $$ Then $$ b_n+\frac{b_{n-1}}{2}=b_{n-1}+\frac{b_{n-2}}{2}=\cdots=b_1+\frac{b_0}{2}=\frac{7}{2}\ln 2, $$ while $$ b_n-b_{n-1}=-\frac{1}{2}(b_{n-1}-b_{n-2})=\cdots=\frac{(-1)^{n-1}}{2^{n-1}}(b_1-b_0)=\frac{(-1)^{n-1}\ln 4}{2^{n-1}}, $$ etc.

Answer 2

$$\ln(a_n)=\frac{\ln(a_{n-1})+\ln(a_{n-2})}{2}$$

This should be a good start !

Answer 3

Do you want to solve $\lim_{n\rightarrow \infty} a_n$?

If so try to compute manually $a_2$, $a_3$, $a_4$ ...

I'm sorry if this is not what are you looking for.