convergence of a fibonacci-like sequence I posted a question earlier on finding a formula for the sequence 
$$t_1, t_2, t_1+t_2, t_1+2t_2,....$$
This is the question I posted earlier
I want to show that as $n\rightarrow \infty$, $\frac{t_{n+1}}{t_n} \rightarrow \phi$
The relationship is $T_n = A_nF_{n-2} +B_n F_{n_1}$
Where $A_1=1, A_2=0, A_{n+2} = A_{n+1}+A_{n}$ and 
$B_1=0, B_2=1, B_{n+2} = B_{n+1}+B_{n}$ 
Since $A_n=F_{n-2}$ and $B_n = F_{n-1}$ then 
$T_n = t_1 F_{n-2}+t_nF_{n-1}$
I'm wondering if this proof will work. I know that this sequence is like Fibonacci and it converges to the golden ratio... 
suppose as $n \rightarrow \infty$, $F_{n+1}/F_n$  converges to a limit $L$. 
Then: $L = \lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n} = \lim_{n\rightarrow \infty} 1 +\frac {1}{L}$
So I solve $L=1+\frac{1}{L} \implies L=\frac{1\pm\sqrt{5}}{2}$
We take the positive root so the answer is $\phi$
The reason i am confused is because this is for convergence $\frac{F_{n+1}}{F_n}$ but my sequence isn't exactly this. 

From linear recurrence theory, as characteristic polynomial of the sequence is $x^2 - x - 1$ and it's roots are $\frac{1 \pm \sqrt{5}}{2}$, we have $t_n = A \left(\frac{1 + \sqrt{5}}{2}\right)^n + B \left(\frac{1 - \sqrt{5}}{2}\right)^n$ where $A$ and $B$ can be found from $t_1$ and $t_2$. If $A \neq 0$ then  $\lim\limits_{n \to \infty} \frac{t_{n + 1}}{t_n} = \lim\limits_{n\to \infty} \left[\frac{1 + \sqrt{5}}{2} + \frac{B}{A}(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})^n\right] / \left[1 + \frac{B}{A}(\frac{1 - \sqrt{5}}{1 + \sqrt{5}})^n\right] = \frac{1 + \sqrt{5}}{2}$ (because $|\frac{1 - \sqrt{5}}{1 + \sqrt{5}}| < 1$), otherwise it is $\frac{1 - \sqrt{5}}{2}$.
And if $A = 0$ then $t_1$ and $t_2$ have different signs - so if they are both positive, $A \neq 0$.
