Fibonacci ratio convergence to golden ratio bound

[billy2908]

Let F_n and F_n+1 be successive Fibonnaci numbers. Then |(F_n+1)/(F_n) - Phi | < 1/(2(F_n)^2)

Where Phi is the Golden ratio.

 

[Eynstone]

Just use the formula for nth fibonacci number.

 

[billy2908]

It's not that simple. I tried fracturing it many times but every time it just seem to leave to the same result.

 

[Matt Benesi]

Remember that the nth Fibonacci number is:

$$F_n = \frac {1} {\sqrt{5}} \left [ \phi^n - \left ( \frac {-1}{\phi}\right)^n\right]$$

and you can multiply F_n² through your equation:

$$\left | F_{n+1} \times F_{n} - \phi \times F^2_n \right | < \frac {1}{2}$$

What happens when you set the whole equation to terms of $\phi$?

Note that for the first 3 Fibonacci numbers: 0,1,1 the formula doesn't work. Additionally, as we get deeper into the Fibonacci sequence, $F_n$ as $n \to \infty$, what does $\left | F_{n+1} \times F_{n} - \phi \times F^2_n \right |$ approach? hint: you might see the value first for a specific very low n