Recurrence relation with limit How do I find the lim for a series that has the following recurrence relation 
$$ a_{0}=2 $$ $$a_1=16 $$ and $$a_{n+1}^2=a_na_{n-1} $$ .I applied $\ln$ and I substituted $$ b_n=\ln (a_n) $$ and so I found the following recurrence relation $$2b_{n+1}=b_n+b_{n-1}$$ .What do I do next? Can someone explain this limit in recurrence relation method because I'm not very familiar with it. I understood that it can only be applied if the given series converges?
 A: So $\forall n \ge 1, b_n - b_{n-1} = c\implies b_n = (b_n - b_{n-1}) + (b_{n-1} - b_{n-2}) + \cdots + (b_1 - b_0) + b_0 = n(b_1 - b_0) + b_0 = nb_1 - (n-1)b_0\implies \ln a_n = n\ln (16) - (n-1)\ln 2 \implies a_n = ...$
A: Hint: 
$$a_{n+1}^2=a_na_{n-1} \;\;\iff\;\; \left(\frac{a_{n+1}}{a_n}\right)^2 = \frac{\;\;\;1\;\;\;}{\cfrac{a_n}{a_{n-1}}}\;\;\iff\;\;\frac{a_{n+1}}{a_n}=\left(\frac{a_{n}}{a_{n-1}}\right)^{-1/2}$$
Iterating all the way down to the beginning of the sequence:
$$
\frac{a_{n+1}}{a_n}=\left(\frac{a_{n}}{a_{n-1}}\right)^{-1/2}=\left(\frac{a_{n-1}}{a_{n-2}}\right)^{(-1/2)^2}= \;\cdots\; = \left(\frac{a_{1}}{a_{0}}\right)^{(-1/2)^n}=8^{(-1/2)^n}
$$
Telescoping:
$$
\begin{align}
a_{n+1} &= 8^{(-1/2)^n} \cdot a_n = 8^{(-1/2)^n+(-1/2)^{n-1}} \cdot a_{n-1} = \cdots = 8^{(-1/2)^n+(-1/2)^{n-1}+\cdots+1} \cdot a_{0} \\[3px]
 &= 8^{\frac{1-(-1/2)^{n+1}}{1-(-1/2)}} \cdot 2 = 8 ^{\frac{2}{3}\left(1-(-1/2)^{n+1}\right)} \cdot 2 = 8 \cdot \left(\frac{1}{4}\right)^{(-1/2)^{n+1}}
\end{align}
$$
In the end $\;\lim_{n \to \infty} a_n = 8 \cdot \left(\cfrac{1}{4}\right)^{\lim_{n \to \infty}(-1/2)^{n}}=\;\cdots$
A: The Ansatz $b_n = x^n$ leads to $2x^2 - x - 1 = 0$ which implies $x = 1,-1/2$. Hence the general solution is $$b_n = c_1 + c_2 \left(-\frac{1}{2} \right)^n.$$
Using the initial conditions we see that $c_1 + c_2 = \ln 2$ and $c_1 - \frac{c_2}{2} = \ln 16$.
