TL;DR

Ratio of terms in the Fibonacci sequence.

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.

From ##a_{n+1} = 1 + a_n^{-1}## and the definition of ##\varphi## we can obtain $$
a_{n+1} - \varphi = (-1)\frac{a_n - \varphi}{a_n\varphi}.$$ Assuming ##a_n \geq 1## we have ##a_n \varphi \geq \varphi > 1## so that $$
|a_{n+1} - \varphi| < |a_n - \varphi|.$$ This is sufficient to conclude that ##a_n \to \varphi##. The factor of ##-1## on the right means that if ##a_n < \varphi## then ##a_{n+1} > \varphi## etc.

There is a much easier way to arrive at these results.

From ##a_{n+1} = 1 + a_n^{-1}## and the definition of ##\varphi## we can obtain $$
a_{n+1} - \varphi = (-1)\frac{a_n - \varphi}{a_n\varphi}.$$ Assuming ##a_n \geq 1## we have ##a_n \varphi \geq \varphi > 1## so that $$

The additional equation that I thought was needed was:
$$a_n - a_{n+2} = \frac{a_n^2 - a_n - 1}{a_n + 1}$$This ties the odd or even sequence to the function values. Because ##x^2 - x - 1## has only one zero on ##(1, 2)##, the only way that ##a_n^2 - a_n - 1## can tend to zero is if ##a_n## tends to ##\varphi##.

From ##a_n \geq 1## we have ##a_n \varphi \geq \varphi ## so that $$
|a_{n+1} - \varphi| \leq \frac{1}{\varphi} |a_n - \varphi|$$ which together with ##\varphi > 1## is sufficient for ##a_n \to \varphi##.

From the last expression for ##F_n## we have $$
\frac{F_{n+1}}{F_n} = \varphi \frac{ A +B \left( - \varphi^{-2} \right)^{n+1} }
{ A + B \left( - \varphi^{-2} \right)^{n} }$$ and some basic analysis shows that as ##n \to \infty## for ##A \neq 0## the fraction tends to 1. It is also easy to see that ##(F_n)## alternates between terms which are greater than and less than ##A\varphi^n##, with ##(F_{n+1}/F_n)## therefore alternatiing between terms which are greater than and less than ##\varphi##.