I have a recursive sequence defined by $$a_{n+1}=\frac{1}{a_n+1}$$, where $a_0=7$.
The sequence is bounded, but it's not monotonic. So how does one show, that the sequence converges? I already found it's limit under the assumption, that it is converging. Which is $a=\frac{\sqrt5-1}{2}$.
Note, that I'm not allowed to use epsilon proofs or proof by induction.
By the recurrence, $$(a_n+1)(a_{n-1}+1)=(\frac{1}{a_{n-1}+1}+1)(a_{n-1}+1)\\=2+a_{n-1}>2$$ since with $a_0=7$ all the terms are obviously positive. Therefore, $$|a_{n+1}-a_n|=\frac{|a_n-a_{n-1}|}{(a_n+1)(a_{n-1}+1)}<\frac12\,|a_n-a_{n-1}|.$$ The rest is a special case of the Banach fixed point theorem: when the distances between consecutive points decrease by a factor less than $1$ the sequence converges (in a complete metric space). As applied to this case, we have by induction: $$|a_{n+1}-a_n|<\frac1{2^n}\,|a_1-a_0|,$$ so the differences decrease at least geometrically. Hence $$|a_{n+m}-a_n|\\\leq (|a_{n+m}-a_{n+m-1}|+\dots+|a_{n+1}-a_n|)\,|a_1-a_0|\\<\left(\frac1{2^{n+m-1}}+\dots+\frac1{2^n}\right)|a_1-a_0|\to0$$ when $n,m\to\infty$ because the geometric series converges. Therefore, $a_n$ is a Cauchy sequence and must converge.
You could even go beyond the problem of the convergence.
This is a first-order rational difference equation (have a look here). If you look at my answer here for a more general case.
Let $m=0$, $x=1$, $y=1$ that is to say $p=-1$ and $q=1$ you should find $$a_n=\frac{K \left(\frac{1}{2} \left(1+\sqrt{5}\right)\right)^n+\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^n}{K \left(\frac{1}{2} \left(\sqrt{5}-1\right)\right)^{-n-1}+\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^{n+1}}$$ and the condition $a_0=7$ leads to $K=\frac{27-7 \sqrt{5}}{22}$ and the limit you already found.
This generates the sequence $$\left\{7,\frac{1}{8},\frac{8}{9},\frac{9}{17},\frac{17}{26},\frac{26}{43},\frac {43}{69},\frac{69}{112},\frac{112}{181},\frac{181}{293},\frac{293}{474},\frac {474}{767},\frac{767}{1241},\frac{1241}{2008},\cdots\right\}$$
the first few values are $$ 7, \frac{1}{8}, \frac{8}{9}, \frac{9}{17} $$ and the last two are both below $1$ and a bit larger than $1/2.$ Then the Mean Value Theorem says, defining $L = \frac{\sqrt 5 - 1}{2},$ that $$ \frac{f(x) - L}{x-L} = f'(\xi) $$ with some $\xi$ between $x$ and $L.$ The derivative is $-1/(1+x)^2 $ and has absolute value below $4/9$ once $ \xi > 1/2.$
Once $1/2 < x < 1 $ we always have $1/2 < x < 1$ So so $$ \left|\frac{f(x) - L}{x-L} \right| < \frac{4}{9} $$ Repeating $f$ says the distance for $L$ is below a power of $4/9$ times the beginning distance.
Write $a_n=L+u_n$, with $L=(\sqrt5-1)/2$, noting that $L^2+L-1=0$. From the recursion we get
$$u_{n+1}={1\over1+L+u_n}-L={1-L-L^2-Lu_n\over1+L+u_n}={-Lu_n\over1+L+u_n}$$
Since $L\approx0.618\lt1$ and $L+u=a_n\gt0$ for all $n$ (by induction: $a_0=7\gt0$, and $a_n\gt0$ implies $a_{n+1}=1/(a_n+1)\gt0$), it follows that
$$|u_{n+1}|={L|u_n|\over|1+L+u_n|}\lt{|u_n|\over|1+L+u_n|}\lt{|u_n|\over|1+0|}=|u_n|$$
Thus $|u_n|$ is a decreasing sequence, hence has a limit (being bounded below by $0$). If that limit were anything other than $0$, we would have
$$\lim_{n\to\infty}|1+L+u_n|=L\lim_{n\to\infty}{|u_n|\over|u_{n+1}|}=L$$
But this is a contradiction, since $|1+L+u_n|=|1+a_n|\gt1\gt L$ for all $n$. Thus $u_n\to0$, so $a_n\to L$ as $n\to\infty$.
Remark: It's easy, but unnecessary, to show that $u_n$ is alternately positive and negative.
