$\{a_n\},\{b_n\},\{c_n\},\{d_n\}$ is series. And $d_n=c_{n-1},c_n=b_{n-1},b_n=a_{n-1},a_n=b_{n-1}+c_{n-1}$
how to proof for any $a_0,b_0,c_0,d_0$ belong to $Z^+$, $\lim\limits_{n\rightarrow \infty} \dfrac{(a_n-a_{n-1})}{a_{n-1}}$ is existent?
In fact,for any $a_0,b_0,c_0,d_0$ belong to $Z-\{0\}$,the result is right.
It is equal to :
for any $a_0,a_1$ belong to $Z-\{0\}$,$a_n=a_{n-2}+a_{n-3}$,$\lim\limits_{n\rightarrow \infty} \dfrac{(a_n-a_{n-1})}{a_{n-1}}$ is existent?
Let the roots of $ x^3 - x - 1 = 0 $ be $ \alpha, \beta, \gamma$. Let $ \alpha$ be the real root. Note that $ |\alpha | > 1 $ and $ |\beta| = |\gamma| < 1$. From the theory of linear recurrences, we get that
$$ a_n = A \alpha^n + B \beta^n + C \gamma^n.$$
Hint: Show that if $ A \neq 0$, we have:
$$ \lim_{n \rightarrow \infty} \frac{ a_n - a_{n-1} } { a_{n-1} } = \lim_{n \rightarrow \infty} \frac{ A \alpha^n - A \alpha^{n-1} } { A \alpha ^{n-1} } = \alpha - 1.$$
Hint: Deal with the case when $A = 0$.
