Exercise 4.4.3 ($\left(\frac{1+\sqrt{5}}{2} \right)^{n-1}

Question

Prove that the Fibonacci numbers satisfiy the inequalities
$$\left(\frac{1+\sqrt{5}}{2} \right)^{n-1} < F_{n+1} < \left(\frac{1+\sqrt{5}}{2} \right)^{n}.$$

Richard Ganaye · Accepted Answer

ewcommand{\N}{\mathbb{N}} \begin{proof} The right inequality is false for $n = 0$, and the left inequality is false for $n = 1$, so we assume that $n\geq 2$. Put $$\lambda = \frac{1-\sqrt{5}}{2},\qquad \mu = \frac{1+\sqrt{5}}{2}.$$ Then $-1<\lambda < 0,\ 1 < \mu$, and by (4.6), $$F_{n+1} = \frac{\mu^{n+1} - \lambda^{n+1}}{\mu - \lambda}.$$ Therefore \begin{align*} F_{n+1} < \left(\frac{1+\sqrt{5}}{2} ight)^{n} & \iff \frac{\mu^{n+1} - \lambda^{n+1}}{\mu - \lambda} < \mu^n\ &\iff \mu^{n+1} - \lambda^{n+1} < \mu^{n+1} - \lambda \mu^n & ( ext{since }\mu - \lambda>0)\ &\iff \lambda \mu^n < \lambda^{n+1}\ &\iff \lambda (\mu^n - \lambda ^n) < 0\ &\iff \lambda F_n < 0 & ( ext{since }\mu - \lambda>0)\ &\iff \lambda <0 &( ext{since }F_n>0 ext{ if } n\geq 1). \end{align*} Since $\lambda < 0$ is true, we obtain $F_{n+1} < \left(\frac{1+\sqrt{5}}{2} ight)^{n}$ if $n>0$. \bigskip Since $\lambda, \mu$ are the roots of the polynomial $x^2 - x - 1$, $\lambda^2 = \lambda +1, \mu^2 = \mu + 1$, thus $ \lambda^2 = \lambda - \mu + \mu^2$. This gives \begin{align*} \left(\frac{1+\sqrt{5}}{2} ight)^{n-1} < F_{n+1} & \iff \mu^{n-1} < \frac{\mu^{n+1} - \lambda^{n+1}}{\mu - \lambda}\ &\iff \mu^{n} - \lambda \mu^{n-1} < \mu^{n+1} - \lambda^{n+1}& (\mu - \lambda>0)\ &\iff \lambda^{n+1} < \mu^{n-1}(\lambda - \mu + \mu^2)\ &\iff \lambda^{n+1} <\mu^{n-1} \lambda^2\ &\iff \left( \frac{\lambda}{\mu} ight)^{n-1} < 1. \end{align*} Since $\left | \frac{\lambda}{\mu} ight | <1$, $\left( \frac{\lambda}{\mu} ight)^{n-1} \leq \left | \frac{\lambda}{\mu} ight |^{n-1} <1$ if $n>1$, thus the preceding inequalities are all true. $$\forall n \in \N,\ n \geq 2 \Rightarrow \left(\frac{1+\sqrt{5}}{2} ight)^{n-1} < F_{n+1} < \left(\frac{1+\sqrt{5}}{2} ight)^{n}.$$ \end{proof}

Exercise 4.4.3 ($\left(\frac{1+\sqrt{5}}{2} \right)^{n-1} < F_{n+1} < \left(\frac{1+\sqrt{5}}{2} \right)^{n}$)

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Exercise 4.4.3 ($\left(\frac{1+\sqrt{5}}{2} \right)^{n-1} < F_{n+1} < \left(\frac{1+\sqrt{5}}{2} \right)^{n}$)

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