Prove that a Fibonacci number is greater than $ φ^n$

Question

How can I prove the following:

If $f_n$ is a number of the Fibonacci sequence and φ= $\frac{1+\sqrt{5}}2$, then $f_n > φ^n$ for every $n >2$?

I have tried using induction but I can't seem to get anywhere.

Answer 1

We have $F_n=\left[ \frac{\phi^n}{\sqrt{5}}\right]$, so this implies $F_n<\phi^n$, see http://en.wikipedia.org/wiki/Fibonacci_number for a proof. So we have the opposite inequality. Here $[x]$ denotes the nearest integer function.

Answer 2

Any solution sequence of the recursion $g_{n+1}=g_n+g_{n-1}$ satisfies $$ g_n=f_{n-1}g_0+f_ng_1 $$ Since $g_n=φ^n$ is a solution of this recursion, $$ φ^n=f_{n-1}φ^0+f_nφ=f_{n-1}+φf_n>f_{n+1}>f_n. $$