Inequality with Fibonacci numbers

The sequence $F_n$ of natural numbers defined by equation $F_{n+2}=F_{n+1}+F_{n}$, with $F_0=0, F_1=1$ is called the Fibonacci sequence. The n-th term in the sequence is called the n-th Fibonacci number.

In many cases, the theory surrounding the Fibonacci numbers would often involve a special number $\phi=\frac{1+\sqrt 5}{2}$ (golden ratio). For every integer $n\geq 1$ is simplest to prove that

$$F_{n+1}<\phi^n$$

Question: do you know of inequalities "closer" of the previous one? I.e., an inequalities

$$F_{n+1}<f(n)$$

such that

$$F_{n+1}<f(n)<\phi^n$$

Thank you very much.

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Well there's a kind of silly derivation: $F_{n+1}=F_{n-1}+F_{n}<\phi^{n-2}+\phi^{n-1}=f(n)<\phi^n$ once you have proven your result by induction. And you can generalize this to get to lower powers of $\phi$. Are you looking for something more 'formulaic' or not as dependent on $\phi$? – Bryan Feb 8 '14 at 9:41
Well, there is an exact expression for $F_n$ in terms of $\phi$ (see the Wikipedia article). That gives a very sharp inequality! – André Nicolas Feb 8 '14 at 9:44
@AndréNicolas, What is it? – Mark Feb 8 '14 at 9:46
@Bryan, I am a curious person. I am looking for any formula! :-) – Mark Feb 8 '14 at 9:47
Andre has told you where to look. – Gerry Myerson Feb 8 '14 at 9:48