How to find the convergence of this sequence? $x_n=1+\frac{1}{x_{n-1}}$ The sequence ($x_n$) is defined by $x_0=1$ and $x_n=1+\frac{1}{x_{n-1}}$ for $n\in ℕ^*$, then how could I find the limit of it? By trying some, I've found that the answer is $\frac{1+\sqrt5}{2}$, but I'm looking for a "legit" way to find it.

In the limit, you have the following continued fraction:
$$x = \lim\limits_{n \to \infty} x_n = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \ddots}}}}.$$
If you're familiar with the golden ratio you'll recognise this is $\varphi$ immediately, but if not, notice that
$$x = 1+\frac{1}{x},$$
which can be rearranged as
$$x^2 -x - 1 = 0.$$
This has roots $$
\frac{1+\sqrt{5}}{2} \;\;\text{ and }\;\; \frac{1-\sqrt{5}}{2}
$$
But since the fraction is clearly positive, $x$ must be the first one.
Note: The limit of your sequence is independent of the value of $x_0$!
