- #1
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- TL;DR
- Ratio of terms in the Fibonacci sequence.
I thought this might be quite interesting to anyone who hasn't seen it before.
Consider any Fibonacci sequence, where ##F_1 > F_0 \ge 0## and ##F_{n+1} = F_n + F_{n-1} \ (n > 1)##. We want to show that the ratio of consecutive terms, ##a_n = \dfrac{F_{n+1}}{F_n}## tends to the Golden Ratio: ##\varphi = \dfrac{1 + \sqrt 5}{2}##.
First, note that ##1 \le a_n \le 2 \ (n \ge 1)##. The sequence is increasing and no term can be more than twice the previous term.
Note that:
$$a_{n + 1} = \frac{F_{n+2}}{F_{n+1}} = \frac{F_{n+1} + F_n}{F_{n+1}} = \frac{(a_n + 1)F_n}{a_nF_n} = \frac{a_n + 1}{a_n} = 1 + \frac 1 {a_n}$$Hence:
$$a_{n+1 } > \varphi \ \Leftrightarrow \ 1 + \frac 1 {a_n} > \varphi \ \Leftrightarrow \ a_n < \frac 1{\varphi - 1}= \varphi$$This last equality follows from the characteristic quadratic equation that defines ##\varphi##:
$$\varphi^2 - \varphi -1 = 0 \ \implies \ \varphi(\varphi-1) = 1 \ \implies \ \varphi = \frac 1 {\varphi - 1}$$So:
If ##a_n < \varphi##, then ##a_{n+1} > \varphi##; and, if ##a_n > \varphi##, then ##a_{n+1} < \varphi##.
This means that ##a_n## is composed of two subsequences, alternating between above and below the Golden Ratio.
Note also that if ##a_n = \varphi##, then ##a_{n+1} = \varphi## and the sequence is constant thereafter. This can only happen, however, when ##F_1 = \varphi F_0##.
If you know a bit of real analysis, you see that it is enough to show that the subsequence less than ##\varphi## is strictly increasing; and, the subsequence above ##\varphi## is strictly decreasing. This guarantees that both subsequences converge to ##\varphi## - as that is the only possible limit.
To see this, we need to compare ##a_{n+2}## with ##a_n##:
$$a_{n+2} = \frac{F_{n+3}}{F_{n+2}} = \frac{F_{n+2} + F_{n+1}}{F_{n+1} + F_n} = \frac{(2a_n + 1)F_n}{(a_n+1)F_n} = \frac{2a_n + 1}{a_n +1}$$Hence:
$$a_n - a_{n+2} = \frac{a_n^2 + a_n - 2a_n - 1}{a_n + 1} = \frac{a_n^2 - a_n - 1}{a_n + 1}$$And we see that:
If ##1 < a_n < \varphi##, then ##a_n^2 - a_n - 1 < 0## and ##1 < a_n < a_{n+2} < \varphi##; and
if ##\varphi < a_n##, then ##a_n^2 - a_n - 1 > 0## and ##\varphi < a_{n+2} < a_n##.
And, as required, we have two monotonic subsequences that must both converge to ##\varphi##.
Note that to prove this last bit rigorously, you need some work, which may not add very much here.
Consider any Fibonacci sequence, where ##F_1 > F_0 \ge 0## and ##F_{n+1} = F_n + F_{n-1} \ (n > 1)##. We want to show that the ratio of consecutive terms, ##a_n = \dfrac{F_{n+1}}{F_n}## tends to the Golden Ratio: ##\varphi = \dfrac{1 + \sqrt 5}{2}##.
First, note that ##1 \le a_n \le 2 \ (n \ge 1)##. The sequence is increasing and no term can be more than twice the previous term.
Note that:
$$a_{n + 1} = \frac{F_{n+2}}{F_{n+1}} = \frac{F_{n+1} + F_n}{F_{n+1}} = \frac{(a_n + 1)F_n}{a_nF_n} = \frac{a_n + 1}{a_n} = 1 + \frac 1 {a_n}$$Hence:
$$a_{n+1 } > \varphi \ \Leftrightarrow \ 1 + \frac 1 {a_n} > \varphi \ \Leftrightarrow \ a_n < \frac 1{\varphi - 1}= \varphi$$This last equality follows from the characteristic quadratic equation that defines ##\varphi##:
$$\varphi^2 - \varphi -1 = 0 \ \implies \ \varphi(\varphi-1) = 1 \ \implies \ \varphi = \frac 1 {\varphi - 1}$$So:
If ##a_n < \varphi##, then ##a_{n+1} > \varphi##; and, if ##a_n > \varphi##, then ##a_{n+1} < \varphi##.
This means that ##a_n## is composed of two subsequences, alternating between above and below the Golden Ratio.
Note also that if ##a_n = \varphi##, then ##a_{n+1} = \varphi## and the sequence is constant thereafter. This can only happen, however, when ##F_1 = \varphi F_0##.
If you know a bit of real analysis, you see that it is enough to show that the subsequence less than ##\varphi## is strictly increasing; and, the subsequence above ##\varphi## is strictly decreasing. This guarantees that both subsequences converge to ##\varphi## - as that is the only possible limit.
To see this, we need to compare ##a_{n+2}## with ##a_n##:
$$a_{n+2} = \frac{F_{n+3}}{F_{n+2}} = \frac{F_{n+2} + F_{n+1}}{F_{n+1} + F_n} = \frac{(2a_n + 1)F_n}{(a_n+1)F_n} = \frac{2a_n + 1}{a_n +1}$$Hence:
$$a_n - a_{n+2} = \frac{a_n^2 + a_n - 2a_n - 1}{a_n + 1} = \frac{a_n^2 - a_n - 1}{a_n + 1}$$And we see that:
If ##1 < a_n < \varphi##, then ##a_n^2 - a_n - 1 < 0## and ##1 < a_n < a_{n+2} < \varphi##; and
if ##\varphi < a_n##, then ##a_n^2 - a_n - 1 > 0## and ##\varphi < a_{n+2} < a_n##.
And, as required, we have two monotonic subsequences that must both converge to ##\varphi##.
Note that to prove this last bit rigorously, you need some work, which may not add very much here.
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