Sequence Of Cube Roots

Let $b_n =\log a_n$ . Then

$\displaystyle b_n=\frac{1}{3}b_{n-1}+\frac{1}{3}b_{n-2}+\frac{1}{3}b_{n-3}, b_0=0, b_1=\log 2, b_2=0$

The characteristic polynomial of the sequence $\{b_n\}$ is

$\displaystyle p(x)=x^3-\frac{1}{3}x^2-\frac{1}{3}x-\frac{1}{3}$

$p(x)$ have three roots; $\displaystyle x=1, -\frac{1}{3}\pm \frac{\sqrt{2}}{3}i$ , so $b_n$ is given as

$\displaystyle b_n=A+Bw_1 ^n+Cw_2 ^n\quad \left(w_1=-\frac{1}{3}+ \frac{\sqrt{2}}{3}i, w_2=-\frac{1}{3}- \frac{\sqrt{2}}{3}i\right)$

for some constant $A, B, C$ .

As $\displaystyle |w_1|=|w_2|=\frac{1}{\sqrt{3}}<1$ , $\displaystyle \lim_{n\to \infty} w_1 ^n=\lim_{n\to\infty} w_2 ^n=0$ . So

$\displaystyle \lim_{n\to\infty} b_n=\lim_{n\to\infty}(A+Bw_1 ^n+Cw_2 ^n)=A$

Besides, the initial condition $b_0=0, b_1=\log 2, b_2=0$ gives a system of equations for $A, B, C$ ; solving this gives $\displaystyle A=\frac{1}{3}\log 2$ .

Therefore

$\displaystyle \lim_{n\to\infty} b_n=\frac{1}{3}\log 2 \quad \therefore \lim_{n\to\infty} a_n=e^{\frac{1}{3} \log 2}=2^{\frac{1}{3}}$

I get the answer, but I don't feel right with my method, please point out for me. $a_{n}=\sqrt[3]{a_{n-1}a_{n-2}a_{n-3}}...(1)$

$a_{n}^3=a_{n-1}a_{n-2}a_{n-3}$

$(a_{n}^3)^3=a_{n-1}^3 a_{n-2}^3 a_{n-3}^3$

$=a_{n-2}a_{n-3}^2a_{n-4}^3a_{n-5}^2a_{n-6}$

$=a_{n-2}a_{n-3}^2a_{n-5}^3a_{n-6}^2a_{n-7}$

...

$=a_{n-2}a_{n-3}^2a_{2}^3a_{1}^2a_{0}$

$\therefore\ a_{n}^3=\sqrt[3]{4a_{n-2}a_{n-3}^2}...(2)$

$\displaystyle \frac{(2)}{(1)}$

$\displaystyle a_{n}^2=\sqrt[3]{ \frac {4a_{n-2}a_{n-3}^2} {a_{n-1}a_{n-2}a_{n-3} }}$

$\displaystyle a_{n}=\sqrt[6]{ \frac {4a_{n-3}} {a_{n-1} }}$

As tends to infinity and limit $a_{n}$ exists (said in the question and that is what we need to find), therefore $a_{n-3}=a_{n-1}$

$\displaystyle lim_{n\to\infty}a_{n}=\sqrt[6]{4}$