Non-linear recurrence with square root: $a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$

Question

How should I approach this problem:

$a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$ where $a_0 = 2, a_1=8$

Answer 1

Hint. One may notice that $a_n>0$, for $n\ge0$, and that $$ \log(a_{n+2})=\frac12\log(a_{n+1})+\frac12\log(a_{n}). $$

Answer 2

An alternative approach to the one given by Oliver Oloa: $$a_{n+2}=\sqrt{a_{n+1}\cdot a_{n}}$$ $$\implies a^2_{n+2}={a_{n+1}\cdot a_{n}}$$ $$\implies \frac{a_{n+2}}{a_{n+1}}=\frac{a_{n}}{a_{n+2}} \tag1$$

Similarly we can write that $$\frac{a_{n+1}}{a_{n}}=\frac{a_{n-1}}{a_{n+1}} \tag2$$ $$\frac{a_{n}}{a_{n-1}}=\frac{a_{n-2}}{a_{n}} \tag3$$ $$\ldots$$ $$\frac{a_{2}}{a_{1}}=\frac{a_{0}}{a_{2}} \tag{n+1}$$

Multiplying these $n+1$ relations, we get $$\frac{a_{n+2}}{a_{n+1}}\cdot\frac{a_{n+1}}{a_{n}}\cdot\frac{a_{n}}{a_{n-1}}\ldots \frac{a_{2}}{a_{1}}=\frac{a_{n}}{a_{n+2}}\cdot\frac{a_{n-1}}{a_{n+1}}\cdot\frac{a_{n-2}}{a_{n}}\ldots\frac{a_{0}}{a_{2}}$$ $$\implies \frac{a_{n+2}}{a_{1}}=\frac{a_{1}\cdot a_{0}}{a_{n+2}\cdot a_{n+1}}$$ $$\implies \frac{a_{n+2}}{8}=\frac{16}{a_{n+2}\cdot a_{n+1}}$$ $$\implies a_{n+2}=\frac{12}{\sqrt{a_{n+1}}}$$ $$\implies a_{n+2}=\sqrt{12}\cdot \sqrt[4]{a_{n}}$$ $$\implies \boxed{a_{n}=\sqrt{12}\cdot \sqrt[4]{a_{n-2}}}$$

So we can write that $$a_{n}=12^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots + \frac{(-1)^{r-1}}{2^{r-1}}}\cdot (a_{n-r})^{\frac{(-1)^r}{2^r}} \,\,\,\,\,\,\, \text{where} \,\,\, r \ge 1$$

For $r=n$, we get $$a_{n}=12^{\frac{1-(-\frac{1}{2})^n}{1-(-\frac{1}{2})}}\cdot (a_0)^{\frac{(-1)^n}{2^n}}$$ $$\boxed{a_{n}=12^{\frac{1}{3}\cdot\frac{2^n-(-1)^n}{2^{n-1}}}\cdot (2)^{\frac{(-1)^n}{2^n}}} \,\,\,\,\,\,\,\,\, \forall \,\,\,\,\,\, n\ge0$$

This is the closed form.

Hope this helps you.