Why does every "fibonacci like" series converge to $\phi$?

Let's define a Fibonacci-like sequence to be a sequence satisfying: $$s_n=s_{n-1}+s_{n-2}.$$ Notice that if we take two Fibonacci-like sequences and add them together, we get another Fibonacci-like sequence - that is, if $a_n$ and $b_n$ are Fibonacci-like, then so is $a_n+b_n$ - you can check the recurrence yourself. Similarly, if we scale every term of a Fibonacci-like sequence by a constant, it is still Fibonacci-like - that is, if $a_n$ is Fibonacci-like, so is $c\cdot a_n$.

This means that such sequences form a structure called a vector space - that is, if we set up a linear combination on Fibonacci-like sequences (e.g. $\alpha\cdot a_n + \beta \cdot b_n$), the result is still Fibonacci-like. However, what's crucial about this insight is that, since we can easily prove that any Fibonacci-like sequence is determined by its first two terms, then if we take two sequences $F_n$, which is the Fibonacci sequence starting with $F_0=0$ and $F_1=1$ and a similar sequence $F'_n=F_{n-1}$, which starts $F'_0=1$ then $F'_1=0$ we can show that if, for a Fibonacci-like sequence $s$ we have $$s_0=s_0\cdot F'_0+s_1\cdot F_0$$ $$s_1=s_0\cdot F'_1+s_1\cdot F_1$$ then, since the right hand side is Fibonacci-like, being a linear combination of Fibonacci-like sequences, and since the right hand sequence agrees for the first two terms, which determine the sequence, we can immediately leap to $$s_n=s_0\cdot F'_n+s_1\cdot F_n$$ or, replacing $F'$ by its definition $$s_n=s_0\cdot F_{n-1}+s_1\cdot{F_n}$$ that is to say, that $s_n$ can be written as a linear combination of the Fibonacci sequence and a shift of it. Then, the ratio of successive terms is $$\frac{s_0\cdot F_{n}+s_1\cdot F_{n+1}}{s_0\cdot F_{n-1}+s_1\cdot F_n}$$ which, if we divide both numerator and denominator by $F_n$ gives $$\frac{s_0+s_1\cdot \frac{F_{n+1}}{F_n}}{s_0\cdot \frac{F_{n-1}}{F_n}+s_1}$$ but since $\frac{F_{n+1}}{F_n}$ tends to $\varphi$ for large $n$ and $\frac{F_{n-1}}{F_n}$ tends to $\varphi^{-1}$, we can see that the above will tend towards $$\frac{s_0+\varphi s_1}{\varphi^{-1} s_0 + s_1}$$ which equals $\varphi$, since the denominator multiplied by $\varphi$ is the numerator. So, this establishes that any Fibonacci-like sequence must have the same limiting ratio as the Fibonacci sequence, which is $\varphi$. A note of caution is that if the denominator, $\varphi^{-1} s_0 + s_1$ is zero, (i.e. if $s_1=\frac{-s_0}{\varphi}$) then this doesn't work - indeed, in this case, it must be that $s_n=s_0\cdot (-\varphi)^{-n}$ and the limiting ratio is $\frac{1}{\varphi}$.

A deeper result is to notice that $s_n=\varphi^n$ and $s'_n=(-\varphi)^{-n}$ are both Fibonacci sequences as well, which lead to Binet's formula when you find a linear combination of those two equalling the Fibonacci sequence. Clearly, the exponentially growing $s_n$ term dominators the decaying $s'_n$ term, which is why $\phi$ is the limit of the ratio of successive Fibonacci numbers; if you want to derive this, notice that $\varphi$ is the solution to $$s_n=s_{n-1}+s_{n-2}$$ $$\varphi^n = \varphi^{n-1} + \varphi^{n-2}$$ Or, dividing by $\varphi^{n-2}$ gives $$\varphi^2=\varphi+1$$ which is indeed a property of the golden ratio, and exactly the equation it comes from.

The sequences that satisfy the linear recurrence

$$G_{n+2} = G_{n+1} + G_n$$

take the general form

$$ A \alpha^n + B \beta^n$$

where $\alpha, \beta$ are roots of

$$x^2 = x + 1 $$

If $A \ne 0$ (and $\alpha$ is the larger root), the ratio of two consecutive such terms $G_{n+1}/G_n$, converges to the larger root, which in this case is the golden ratio: $\varphi$.

$$ G_{n+1}/G_n = \frac{A \alpha^{n+1} + B \beta^{n+1}}{A \alpha^n + B \beta^n}$$

Take the larger root out ($\alpha$)

$$ = \alpha\left(\frac{A + B (\beta/\alpha)^{n+1}}{A + B (\beta/\alpha)^n}\right) \to \alpha$$

If $A = 0$ and $B \ne 0$, then it converges to $\beta$.

In case of interest, this phenomenon is not restricted to Fibonacci and Lucas numbers, or that law. Even if we stick with linear degree two recurrences, for example $$ \color{magenta}{ x_{n+2} = 4 x_{n+1} - x_n}, $$ the ratio of consecutive numbers in the sequence has a well defined limit, in this case the largest root of $$ \lambda^2 - 4 \lambda + 1 = 0, $$ or $$ 2 + \sqrt 3 \approx 3.73205, $$ at least as long as we start with, say, $x_1 \geq x_0.$ So, take $$ 1, 1, 3, 11, 41, 153, 571, 2131, 7953, 29681, $$ or $$ 1, 2, 7, 26, 97, 362, 1351, 5042, 18817, $$ For the first sequence, there are constants $A,B,$ with $A > 0,$ such that $$ x_n = A \, \left(2 + \sqrt 3 \right)^n + B \, \left(2 - \sqrt 3 \right)^n. $$ For the second sequence, there are (probably different) constants $C,D,$ with $C > 0,$ such that $$ x_n = C \, \left(2 + \sqrt 3 \right)^n + D \, \left(2 - \sqrt 3 \right)^n. $$

For either sequence, a number divided by the previous number in the sequence is approximately $2 + \sqrt 3.$

$S_n=\phi^n$ and $S_n=(-\phi)^n$ are two solutions of the Fibonacci recurrence $S_n+S_{n+1}=S_{n+2}$, because $1+\phi=\phi^2$ and $1-\phi^{-1}=\phi^{-2}$.

By linearity, any combination $S_n=a\phi^n+b(-\phi)^{-n}$ is also a solution. You can adjust $a$ and $b$ to match any two initial values. For instance,

$$S_0=3=a\cdot1+b.1,\\S_1=4=a\cdot\phi-b.\phi^{-1},$$ giving $$a=\frac{3\phi^{-1}+4}{\phi+\phi^{-1}},\\b=\frac{3\phi-4}{\phi+\phi^{-1}}.$$ For growing $n$ the powers of $-\phi$ quickly become neglictible (except if $a=0$), and $$S_n\approx a\cdot\phi^n,$$a geometric progression of ratio $\phi$.

$$\begin{align} & S_n & a\phi^n\ \ \ \ \ \ \ \ \\ & 3 & 2.61803398875 \\ & 4 & 4.23606797750 \\ & 7 & 6.85410196625 \\ & 11 & 11.0901699437 \\ & 18 & 17.9442719100 \\ & 29 & 29.0344418537 \\ & 47 & 46.9787137637 \\ & 76 & 76.0131556175 \\ & 123 & 122.991869381 \\ & 199 & 199.005024999 \\ & 322 & 321.996894380 \\ & 521 & 521.001919379 \\ & 843 & 842.998813759 \\ & 1364 & 1364.00073314 \\ & 2207 & 2206.99954690 \\ & 3571 & 3571.00028003 \\ & 5778 & 5777.99982693 \\ & 9349 & 9349.00010696 \\ & 15127 & 15126.9999339 \\ & 24476 & 24476.0000409 \\ \end{align}$$