Question

If F_n is the n-th Fibonacci number, show that lim $\lim_{n \to \infty} \frac { F_{n+1}}{F_n} = \frac{1+\sqrt{5}}{2}$

Answer 1

Prove that

$$F_n=\dfrac{1}{\sqrt{5}}\bigg((\dfrac{1+\sqrt{5}}{2})^{n}-(\dfrac{1-\sqrt{5}}{2})^{n}\bigg)$$

Let’s  look at the interesting quadratic

$$x^2-x-1=0$$

The roots of this quadratic are

$$x=\dfrac{1\pm\sqrt{5}}{2}$$

This quadratic can also be written as

$$x^2=x+1$$

From this, we can write expressions for all $x^n:$

$$x=x$$

$$x^2=x+1$$

$$x^3=x\cdot x^2=x(x+1)$$ $$=x^2+x=x+1+x=2x+1$$

$$x^4=x\cdot x^3=x(2x+1)$$ $$=2x^2+x=2x+2+x=3x+2$$

$$x^5=5x+3=x^3+x^4$$

$$x^6=8x+5=x^4+x^5$$

$$...$$

We note that

$$x^n=F_nx+F_{n-1}$$

Let the roots of our original quadratic be $u=\dfrac{1+\sqrt{5}}{2},$ and $v=\dfrac{1-\sqrt{5}}{2}.$

Since both $u$ and $v$ are roots of the quadratic, they must both satisfy 

$$x^n=F_nx+F_{n-1}$$

 So

$$u^n=F_nu+F_{n-1}$$

$$v^n=F_nv+F_{n-1}$$

Subtracting the second equation from the first equation yields

$$u^n-v^n=F_n(u-v)$$

Then

$$(\dfrac{1+\sqrt{5}}{2})^{n+1}-(\dfrac{1-\sqrt{5}}{2})^{n+1}$$

$$=F_n(\dfrac{1+\sqrt{5}}{2}-\dfrac{1-\sqrt{5}}{2})$$

$$=\sqrt{5}F_n$$

Then we have

$$F_n=\dfrac{1}{\sqrt{5}}\bigg((\dfrac{1+\sqrt{5}}{2})^{n}-(\dfrac{1-\sqrt{5}}{2})^{n}\bigg)$$

$$F_{n-1}=\dfrac{1}{\sqrt{5}}\bigg((\dfrac{1+\sqrt{5}}{2})^{n-1}-(\dfrac{1-\sqrt{5}}{2})^{n-1}\bigg)$$

$$\dfrac{F_n}{F_{n-1}}=\dfrac{(\dfrac{1+\sqrt{5}}{2})^{n}-(\dfrac{1-\sqrt{5}}{2})^{n}}{(\dfrac{1+\sqrt{5}}{2})^{n-1}-(\dfrac{1-\sqrt{5}}{2})^{n-1}}$$

$$=\dfrac{\dfrac{1+\sqrt{5}}{2}-\dfrac{1-\sqrt{5}}{2}\cdot(\dfrac{1-\sqrt{5}}{1+\sqrt{5}})^{n-1}}{1-(\dfrac{1-\sqrt{5}}{1+\sqrt{5}})^{n-1}}$$

$$\lim\limits_{n\to \infin}\dfrac{F_n}{F_{n-1}}=$$

$$=\lim\limits_{n\to \infin}\dfrac{\dfrac{1+\sqrt{5}}{2}-\dfrac{1-\sqrt{5}}{2}\cdot(\dfrac{1-\sqrt{5}}{1+\sqrt{5}})^{n-1}}{1-(\dfrac{1-\sqrt{5}}{1+\sqrt{5}})^{n-1}}$$

$$=\dfrac{\dfrac{1+\sqrt{5}}{2}-\dfrac{1-\sqrt{5}}{2}\cdot0}{1-0}$$

$$=\dfrac{1+\sqrt{5}}{2}$$