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. 
 A: What you are looking for is practically a Fibonacci recurrence,  with starting 
values different from $F_1=1,\; F_2=1$, like it is for Lucas sequence .
Clearly, if your $T_1 , T_2$ happens to be equal to  two consecutive F's, then you just have
a shifted Fibonacci Sequence.
Also in your case we have the matrix relation
$$
\left( {\matrix{   {T_{\,k + 2} }  \cr    {T_{\,k + 1} }  \cr 
 } } \right) = \left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)\left( {\matrix{   {T_{\,k + 1} }  \cr    {T_{\,k} }  \cr 
 } } \right)\quad  \Rightarrow \quad \left( {\matrix{   {T_{n + 2} }  \cr    {T_{\,n + 1} }  \cr 
 } } \right) = \left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)^{\,n} \left( {\matrix{   {T_{\,2} }  \cr    {T_{\,1} }  \cr 
 } } \right)
$$
The  matrix elevated to $n$ can also be written as
$$
\left( {\matrix{
   1 & 1  \cr 
   1 & 0  \cr 
 } } \right)^{\,n}  = \left( {\matrix{
   {F_{\,n + 1} } & {F_{\,n} }  \cr 
   {F_{\,n} } & {F_{\,n - 1} }  \cr 
 } } \right)
$$
therefore
$$
{{T_{n + 2} } \over {T_{\,n + 1} }} = {{T_{\,2} F_{\,n + 1}  + T_{\,1} F_{\,n} } \over {T_{\,2} F_{\,n}  + T_{\,1} F_{\,n - 1} }}
 = {{\left( {T_{\,2}  + T_{\,1} } \right)F_{\,n}  + T_{\,2} F_{\,n - 1} } \over {T_{\,2} F_{\,n}  + T_{\,1} F_{\,n - 1} }}
 = {{\left( {T_{\,2}  + T_{\,1} } \right)F_{\,n} /F_{\,n - 1}  + T_{\,2} } \over {T_{\,2} F_{\,n} /F_{\,n - 1}  + T_{\,1} }}
$$
and the limit follows easily
Also refer to the para. "Closed-form expression" in the Wikipedia article
--- answer to your comment  ---
The vectorial representation is just a translation of the recursive identity (plus an obvious one)
$$
\left\{ \matrix{
  T_{\,n + 2}  = T_{\,n + 1}  + T_{\,n}  \hfill \cr 
  T_{\,n + 1}  = T_{\,n + 1}  \hfill \cr}  \right.\quad  \Rightarrow \quad 
\left( {\matrix{   {T_{\,n + 2} }  \cr    {T_{\,n + 1} }  \cr  } } \right)
 = \left( {\matrix{   1 & 1  \cr    1 & 0  \cr  } } \right)
\left( {\matrix{   {T_{\,n + 1} }  \cr    {T_{\,n} }  \cr  } } \right)
$$
which has the advantage that multiple recursion steps are translated into matrix multiplication (power)
$$
\eqalign{
  & \left( {\matrix{   {T_{\,n + 2} }  \cr    {T_{\,n + 1} }  \cr 
 } } \right) = \left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)\left( {\matrix{   {T_{\,n + 1} }  \cr    {T_{\,n} }  \cr 
 } } \right) = \left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)\left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)\left( {\matrix{   {T_{\,n} }  \cr    {T_{\,n - 1} }  \cr 
 } } \right) =  \cdots  =   \cr   &  = \left( {\matrix{   1 & 1  \cr  1 & 0 \cr
 } } \right)^{\,n} \left( {\matrix{   {T_{\,2} }  \cr    {T_{\,1} }  \cr 
 } } \right) \cr} 
$$
Since the Fibonacci N. obey to the same recurrence
$$
\eqalign{
  & \left( {\matrix{   {F_{\,n + 2} }  \cr    {F_{\,n + 1} }  \cr 
 } } \right) = \left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)^{\,n} \left( {\matrix{   {F_{\,2} }  \cr    {F_{\,1} }  \cr 
 } } \right) = \left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)^{\,n} \left( {\matrix{   1  \cr    1  \cr 
 } } \right)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)^{\,n}  = \left( {\matrix{   {F_{\,n + 1} } & {F_{\,n} }  \cr    {F_{\,n} } & {F_{\,n - 1} }  \cr 
 } } \right) \cr} 
$$
and
$$
\eqalign{
  & \left( {\matrix{   {T_{n + 2} }  \cr    {T_{\,n + 1} }  \cr 
 } } \right) = \left( {\matrix{   1 & 1  \cr    1 & 0  \cr 
 } } \right)^{\,n} \left( {\matrix{   {T_{\,2} }  \cr    {T_{\,1} }  \cr 
 } } \right) = \left( {\matrix{   {F_{\,n + 1} } & {F_{\,n} }  \cr    {F_{\,n} } & {F_{\,n - 1} }  \cr 
 } } \right)\left( {\matrix{   {T_{\,2} }  \cr    {T_{\,1} }  \cr 
 } } \right)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  T_{\,n + 2}  = T_{\,2} F_{\,n + 1}  + T_{\,1} F_{\,n}  = T_{\,2} F_{\,n}  + T_{\,2} F_{\,n - 1}  + T_{\,1} F_{\,n}  \hfill \cr 
  T_{\,n + 1}  = T_{\,2} F_{\,n}  + T_{\,1} F_{\,n - 1}  \hfill \cr}  \right. \cr} 
$$
A: 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, $A \neq 0$.
