How to find the convergence of this sequence? $x_n=1+\frac{1}{x_{n-1}}$

Question

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.

Answer 1

If the limit exists, then let's call it $x$.

Then, $$x = \lim_{n\to\infty} x_n = \lim_{n\to\infty} \left(1 + \frac{1}{x_{n-1}}\right) = 1 + \frac{1}{\lim_{n\to\infty} x_{n-1}} = 1 + \frac{1}{x}$$ and $$x^2-x-1=0$$

So there are only two possible limits.

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You can then show that the sequence is positive, which leaves only one possible limit.

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All that is left then is to show that the sequence actually has a limit.

Answer 2

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 $x_0$!