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.
How to find the convergence of this sequence? $x_n=1+\frac{1}{x_{n-1}}$
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calculus
sequences-and-series
recurrence-relations
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0You may want to look up the "Fibonacci sequence" and find the relation with your sequence. – 2017-01-04
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0You can probably find some other posts about the same sequence. For searching using a mathematical formula, you might [try Approach0](https://approach0.xyz/search/?q=%24x_n%3D1%2B%5Cfrac%7B1%7D%7Bx_%7Bn-1%7D%7D%24&p=1). – 2017-01-04
2 Answers
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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.
You can then show that the sequence is positive, which leaves only one possible limit.
All that is left then is to show that the sequence actually has a limit.
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0@JohnAdams Yes. But keep in mind, that only shows that if the sequence is convergent, then the limit is whatever you calculated. You still have to show the sequence converges. – 2017-01-04
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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 $x_0$!