Assuming

- $F_n = F_{n-1} + F_{n-2}$
- $F_n > F_{n-1}$
- $F_{n-1} > F_{n-2}$

$=>$

Start with $F_n = F_{n-1} + F_{n-2}$, and make the right side bigger by replacing $F_{n-2}$ with $F_{n-1}$

$F_n < F_{n-1} + F_{n-1}$

$=>$

$F_n < 2F_{n-1}$

$=>$

$F_n < 2^n$

So the fibonacci sequence, one item at a time, grows more slowly than $2^n$.

But on the other hand every 2 items the Fibonacci sequence more than doubles itself:

$(1) F_n = F_{n-1} + F_{n-2}$

$(2) F_{n-1} = F_{n-2} + F_{n-3}$

$=>$

Replace $F_{n-1}$ in $(1)$ with $F_{n-1}$ from $(2)$

$F_n = 2F_{n-2} + F_{n-3}$

$=>$

Because $F_{n-3}$ is greater than zero, we can drop it from the right side, and that makes the right side smaller than the left.

$F_n > 2F_{n-2}$

$=>$

$F_n > 2^{n/2}$

$F_n > (2^{1/2})^n$

$F_n > \sqrt{2}^n$

Because the Fibonacci sequence is bounded between two exponential functions, it's effectively an exponential function with the base somewhere between `1.41`

and `2`

.

${1.41}^n$ < $F_n < 2^n$

That base ends up being the golden ratio. See https://math.stackexchange.com/a/1201069/62698