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How can I prove the following:

If $f_n$ is a number of the Fibonacci sequence and φ= $\frac{1+\sqrt{5}}2$, then $f_n > φ^n$ for every $n >2$?

I have tried using induction but I can't seem to get anywhere.

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    $\begingroup$ $\varphi^1 > 1 = f_1$ $\endgroup$
    – GEdgar
    Mar 1, 2014 at 19:26
  • $\begingroup$ Sorry, I forgot to say that the affirmation is valid for $n>2$. $\endgroup$
    – Lstoi
    Mar 1, 2014 at 23:42
  • $\begingroup$ Then take $n=3$. We have $f_3=2$, but $\phi^3\sim 4.236$, so the claim is not true. In fact, the opposite inequality holds. $\endgroup$ Mar 2, 2014 at 19:00
  • $\begingroup$ See math.stackexchange.com/a/1343414/589 $\endgroup$
    – lhf
    Jan 12, 2021 at 9:41

2 Answers 2

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We have $F_n=\left[ \frac{\phi^n}{\sqrt{5}}\right]$, so this implies $F_n<\phi^n$, see http://en.wikipedia.org/wiki/Fibonacci_number for a proof. So we have the opposite inequality. Here $[x]$ denotes the nearest integer function.

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  • $\begingroup$ This is not true as written. For example, $F_3=2>1.89\dots=\phi^3/\sqrt5$. We have $F_n=(\phi^n-(1-\phi)^n)/\sqrt5$, so $F_n$ is larger than $\phi^n/\sqrt 5$ whenever $n$ is odd. What you get is that $F_n$ is the integer closest to $\phi^n/\sqrt5$, and a little additional argument is needed to see that this still gives $F_n<\phi^n$. $\endgroup$ Mar 2, 2014 at 18:14
  • $\begingroup$ Sorry, the brackets denote nearest integer function. So $F_3=[1.89]=2<\phi^3$ is correct. But thank you, I corrected the brackets. $\endgroup$ Mar 2, 2014 at 18:51
  • $\begingroup$ Ah, OK. Had not seen the notation used that way, seems unnecessarily ambiguous. Thanks for clarifying. $\endgroup$ Mar 2, 2014 at 19:54
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Any solution sequence of the recursion $g_{n+1}=g_n+g_{n-1}$ satisfies $$ g_n=f_{n-1}g_0+f_ng_1 $$ Since $g_n=φ^n$ is a solution of this recursion, $$ φ^n=f_{n-1}φ^0+f_nφ=f_{n-1}+φf_n>f_{n+1}>f_n. $$

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