Ratio of terms in sequences satisfying $a_{n}+a_{n+1}=a_{n+2}$ - always tends to the golden ratio In the Fibonacci sequence ($f_{0}=0,f_{1}=1,f_{n}=f_{n-1}+f_{n-2}$), it has been proven that
$$\lim_{n\rightarrow \infty}\frac{f_{n+1}}{f_{n}}=\varphi$$
In the Lucas sequence ($L_{0}=2,L_{1}=1,L_{n}=L_{n-1}+L_{n-2}$), it has similarly been proven that
$$\lim_{n\rightarrow \infty}\frac{L_{n+1}}{L_{n}}=\varphi$$
I've been looking into the Tertiary Fibonacci series ($q_{0}=3,q_{1}=2,q_{n}=q_{n-1}+q_{n-2}$) and, although I haven't proven it, I've seen that the ratios of consecutive Tertiary Fibonacci terms also tends to the golden ratio.
Is it typically, or always, the case that, for sequences satisfying
$t_{0}=a, t_{1}=b (a>0 \wedge b>0)$ and $t_{n}=t_{n-1}+t_{n-2}$,
$$\lim_{n\rightarrow \infty}\frac{t_{n+1}}{t_{n}}=\varphi?$$
rtybase suggested that this is similar to another query, but I do not understand the solution to that one or even the question. The answers provided for this one have been a lot easier to understand, not to mention more relevant to the topic.
 A: It can be shown that any such sequence $t_n$ has a formula of the form $t_n=A\varphi^n+B\left(-\frac{1}{\varphi}\right)^{n}$. If $A \ne 0$, then it follows immediately from this formula that $\lim_{n \to \infty} \frac{t_{n+1}}{t_n}=\varphi$, because $|\varphi|>\left|-\frac{1}{\varphi}\right|$.
However, if $A=0$, then we have $t_n=B\left(-\frac{1}{\varphi}\right)^n$. In this case $t_1=-\frac{1}{\varphi}t_0$ and so it is not possible to have both $t_1>0$ and $t_0>0$.
In summary, the answer is yes: any sequence of the given form must have the desired limit. However the condition that $t_1$ and $t_0$ are both positive is crucial. Otherwise the sequence $t_n=\left(-\frac{1}{\phi}\right)^n$ would be a counterexample.
A: Note: In this approach, I have assumed that the limit exists now, we show that it equals the golden ratio.
Suppose we have $\displaystyle\lim_{n \to \infty}{\frac{t_{n+1}}{t_n}}=q$.
As, $$t_{n+1}=t_n+t_{n-1}$$ we divide both sides of this equation by $t_n$ and impose the limit both sides.
Now as $$n \to \infty$$
Hence we have $$\lim_{n\to \infty}\frac{t_{n+1}}{t_n}= \lim_{n\to \infty}\frac{t_{n}}{t_{n-1}}=q$$.
Then, $$q=\frac{1}{q}+1$$ from which we observe that $q$ is the golden ratio irrespective of values of $a$ and $b$! The other root cannot be considered here as it is negative and the limit is always positive.
Proving that the limit exists:
Let $$\frac{t_{n+1}}{t_n}= A_{n+1}$$.
Now,  we have $$A_{n+1}=1+\frac{1}{A_n}$$.
If the limit is non existent, then $A_n$ must tend to infinity.
But this would mean that $A_{n+1} $ would tend to $1$. This is a contradiction as $A_{n+1}$ should also have tended to infinity. Hence the limit must exist.
A: Doing the calculations I have verified that for a_0 non-zero and a_1=a_0/Fi the limit of the quotient is
sqrt(5)*Fi-4 which is different from Fi
