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Let $\{f_n\}_{(n\ge 1)}$ Fibonacci sequence, which satisfies the following recurrence relation; $$f_1=f_2=1, f_{n+2}=f_n+f_{n+1} \phantom{a} (n\ge 1)$$ A few months ago I found an interesting inequality; $$\forall N\in\mathbb{N};\phantom{a}\sum_{n=1}^{N}\frac{f_n}{f_{n+1}}>\frac{-1+\sqrt{5}}{2}N\quad\cdots (1)$$ This can be proved amazingly simple;

Let $\displaystyle S_N=\sum_{n=1}^{N}\frac{f_n}{f_{n+1}}$ and $\displaystyle T_N=\sum_{n=1}^{N}\frac{f_{n+1}}{f_n}$. By Cauchy-Schwarz inequality $S_N T_N> N^2$. Using the given recurrence relation we can easily prove that $T_N<S_N+N$. Combining two results, $(N+S_N)S_N>T_N S_N>N^2$, so $S_N^2+NS_N-N^2>0$. Solving this quadratic inequality, we get $\displaystyle S_N>\frac{-1+\sqrt{5}}{2}N$.

After proving (1) I tried to find the upper bound of the sum by using similar method. I finally found $$\forall N\in\mathbb{N}; \phantom{a}\sum_{n=1}^{N}\frac{f_{n+1}}{f_n}<\frac{1+\sqrt{5}}{2}N+\sum_{n=1}^{N}\frac{1}{n}\quad\cdots (2)$$ However, my proof of (2) is long and complex than I expected; Is there any method to obtain (better) upper bound by using simple method like the proof of (1)? Let's share some ideas.

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up vote 4 down vote accepted

HINT A stronger statement is in fact true. $$\dfrac{f_{n+1}}{f_n} < \dfrac{1+\sqrt{5}}2 + \dfrac1n$$ and $$\dfrac{f_{n}}{f_{n+1}} < \dfrac{\sqrt{5}-1}2 + \dfrac1{n+1}$$

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Thanks for comment. I actually used $\displaystyle \frac{f_{n+1}}{f_n}<\frac{1+\sqrt{5}}{2}+\frac{1}{n}$ to prove (2). To prove this inequality, I used Binet's formula; $\displaystyle f_n=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n \sqrt{5}}$. Are there some other methods to prove this? – hunminpark Jan 10 '13 at 3:11

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