## Task 5 - 301235

Investigate whether the sequence $\left(x_{n}\right)$ defined by

$$
x_{1}=1, \quad x_{n+1}=\frac{1}{x_{n}+1} \quad(n=1,2,3, \ldots)
$$

is convergent, and determine its limit if it is convergent.
First, the calculations here will be somewhat simpler if we also include the initial term $x_{0}=0$, which is obviously compatible with the above recursion rule. Since $x_{0}$ and each $x_{n}(n \in \mathbb{N})$ is rational, $x_{n+1}$ is also rational, so we can work with the approach

$$
x_{n}=\frac{a_{n}}{b_{n}} \quad(n \in \mathbb{N})
$$

where $\left(a_{n}\right)$ and $\left(b_{n}\right)$ are two sequences of natural numbers with $\operatorname{gcd}\left(a_{n}, b_{n}\right)=1$ and the following equations hold

$$
\frac{a_{0}}{b_{0}}=\frac{0}{1}, \quad \frac{a_{1}}{b_{1}}=\frac{1}{1}, \quad \frac{a_{n+1}}{b_{n+1}}=\frac{1}{\frac{a_{n}}{b_{n}}+1}=\frac{b_{n}}{a_{n}+b_{n}}
$$

from which we can immediately derive the simple recursive relationship

$$
a_{0}=0, \quad a_{1}=1, \quad a_{n+2}=b_{n+1}=b_{n}+a_{n}=a_{n+1}+a_{n} \quad(n \in \mathbb{N})
$$

The sequence $\left(b_{n}\right)$ is then, due to

$$
b_{n}=a_{n+1} \quad(n \in \mathbb{N})
$$

very similar to the sequence $\left(a_{n}\right)$ and is simply shifted one position to the right relative to it.

Next, we determine an explicit formula for the sequence $\left(a_{n}\right)$ from the recursion $\left(^{*}\right)$ using the usual approach

$$
a_{n}=c_{1} q_{1}^{n}+c_{2} q_{2}^{n}
$$

where $q_{1}$ and $q_{2}$ are the two solutions of the equation $q^{2}-q-1=0$, which are then

$$
q_{1,2}=\frac{1 \pm \sqrt{5}}{2}
$$

and the values of $c_{1}$ and $c_{2}$ are determined from

$$
c_{1} q_{1}^{0}+c_{2} q_{2}^{0}=a_{0}=0, \quad c_{1} q_{1}+c_{2} q_{2}=a_{1}=1
$$

to

$$
c_{1}=-c_{2}=\frac{1}{\sqrt{5}}
$$

which ultimately results in

$$
a_{n}=\frac{1}{\sqrt{5}}\left(q_{1}^{n}-q_{2}^{n}\right), \quad b_{n}=a_{n+1}=\frac{1}{\sqrt{5}}\left(q_{1}^{n+1}-q_{2}^{n+1}\right) \Rightarrow x_{n}=\frac{q_{1}^{n}-q_{2}^{n}}{q_{1}^{n+1}-q_{2}^{n+1}} \quad(n \in \mathbb{N})
$$

In particular, it follows from this that because $q_{2} \approx-0.61$, $\left(q_{2}^{n}\right)$ is a null sequence, and thus the sequence $\left(x_{n}\right)$ converges, and with the limit

$$
\lim _{n \rightarrow \infty} x_{n}=\frac{1}{q_{1}}=\frac{\sqrt{5}-1}{2}
$$

which answers all the questions here.