# How to prove that $\mathrm{Fibonacci}(n) \leq n!$, for $n\geq 0$

I am trying to prove it by induction, but I'm stuck $$\mathrm{fib}(0) = 0 < 0! = 1;$$ $$\mathrm{fib}(1) = 1 = 1! = 1;$$

Base case n = 2,

$$\mathrm{fib}(2) = 1 < 2! = 2;$$

Inductive case assume that it is true for (k+1) $k$ Try to prove that $\mathrm{fib}(k+1) \leq(k+1)!$

$$\mathrm{fib}(k+1) = \mathrm{fib}(k) + \mathrm{fib}(k-1) \qquad(LHS)$$

$$(k+1)! = (k+1) \times k \times (k-1) \times \cdots \times 1 = (k+1) \times k! \qquad(RHS)$$

......

How to prove it?

• By the way, you have to assume it is true for $k$, not for $k+1$. Otherwise, you are simply assuming what you want to prove Jun 6, 2012 at 17:16
• Base case is $0$, $1$ (to allow use of the Fibonacci recursion) Jun 6, 2012 at 17:18
• Not that it matters for your problem, but as Ayman observed below, n! is a very loose upper bound. Jun 6, 2012 at 17:54
• The bound is ridiculously bad, why do you need that ? Jun 15, 2012 at 17:38

$$F_{k+1} = F_k + F_{k-1} \le k! + (k - 1)! \le k! + k! \le 2 k! \le (k + 1) k!$$
• Given that $F_k \sim \varphi^k$, it's easy to see that $k!$ grows much faster than $F_k$. Jun 6, 2012 at 17:19
• Short and sweet. (Though you can make it even shorter: $k!+(k-1)!=(k+1)(k-1)!\le(k+1)!$.) Jun 6, 2012 at 22:11