Proving that $\alpha^{n-2}\leq F_n\leq \alpha^{n-1}$ for all $n\geq 1$, where $\alpha$ is the golden ratio

Question

I got stuck on this exercise. It is Theorem 1.15 on page 14 of Robbins' Beginning Number Theory, 2nd edition.

Theorem 1.15. $\alpha^{n-2}\leq F_n\leq \alpha^{n-1}$ for all $n\geq 1$.

Proof: Exercise.

(image of relevant page)

Answer 1

If you check the inequality when $n=1$ and $n=2$, you can argue inductively, since the powers of $\alpha$ satisfy the same order two recursion as do the fibonacci numbers.

That is, assume the statement holds for each of $n$ and $n+1$, then just add up the inequalities; you'll use e.g. $\alpha^{n-2}+\alpha^{n-1}=\alpha^n$, which follows from the relation $1+\alpha=\alpha^2$ on multiplying by $\alpha^{n-2}$. The right sides work the same way.

Answer 2

I'll give you a hint (two, actually): use strong induction, and note that

$$ \alpha^2 = \alpha + 1$$

See if you can get somewhere from there :)

Answer 3

Okay, we want to prove that $$\alpha^{n-2}\leq F_n \leq \alpha^{n-1}.$$ For this all we need is weak mathematical induction. For our base case, pick $n=1$, and we have $$\alpha^{-1}=\frac{2}{1+\sqrt{5}}< 1 \leq \alpha^0=1.$$ Check. Now let's proceed with the inductive step. By inductive hypothesis, $$\alpha^{n-3}\leq F_{n-1} \leq \alpha^{n-2}$$ and $$\alpha^{n-4}\leq F_{n-2} \leq \alpha^{n-3}$$ so, since $F_{n}=F_{n-1}+F_{n-2}$, $$\alpha^{n-3}+\alpha^{n-4}\leq F_{n} \leq \alpha^{n-2}+\alpha^{n-3}$$ So we see it suffices to show that $$\alpha^{m-2}+\alpha^{m-1}=\alpha^{m}.$$ To prove this part, let's do a little subinduction on $m$. (This could just have easily been proven in a lemma.) Note that it is indeed true that $$\alpha^2=\frac{3+ \sqrt{5}}{2}=\alpha+1.$$ So now, by induction on $m$, $$\begin{eqnarray*}\alpha^{m-1}+\alpha^{m}&=&\alpha\left(\alpha^{m-2}+\alpha^{m-1}\right)\\&=&\alpha(\alpha^m)=\alpha^{m+1}.\end{eqnarray*}$$ And we're done.

Answer 4

Using Binet's Fibonacci Number Formula,

$\alpha+\beta=1,\alpha\beta=-1$ and $\beta<0$ \begin{align} F_n-\alpha^{n-1}= & \frac{\alpha^n-\beta^n}{\alpha-\beta}-\alpha^{n-1} \\ = & \frac{\alpha^n-\beta^n-(\alpha-\beta)\alpha^{n-1}}{\alpha-\beta} \\ = & \beta\frac{(\alpha^{n-1}-\beta^{n-1})}{\alpha-\beta} \\ = & \beta\cdot F_{n-1}\le 0 \end{align} if $n\ge 1$.

\begin{align} F_n-\alpha^{n-1}= & \frac{\alpha^n-\beta^n}{\alpha-\beta}-\alpha^{n-2} \\ = & \frac{\alpha^{n-1}\cdot \alpha-\beta^n-\alpha^{n-1}+\beta\cdot \alpha^{n-2}}{\alpha-\beta} \\ = & \frac{\alpha^{n-1}\cdot (1-\beta)-\beta^n-\alpha^{n-1}+\beta\cdot \alpha^{n-2}}{\alpha-\beta} \\ = & \frac{\beta(\alpha^{n-2}-\alpha^{n-1})-\beta^n}{\alpha-\beta} \\ = &\frac{\beta\cdot \alpha^{n-2}(1-\alpha)-\beta^n}{\alpha-\beta} \\ = & \frac{\beta\cdot \alpha^{n-2}(\beta)-\beta^n}{\alpha-\beta} \\ = & \beta^2\frac{\alpha^{n-2}-\beta^{n-2}}{\alpha-\beta} \\ = & \beta^2F_{n-2}\ge 0 \end{align} if $n\ge 1$.