I Useless continued fraction for 1

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Does continued fraction representing 1 converge?
I'm interested to know whether the equation

$$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$

is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
 
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Say RHS continued fraction is x, we have an equation
$$x=2-\frac{1}{x}$$
$$(x-1)^2=0$$
$$x=1$$
So we know if x converges, x=1. If it does not converge the equation is not satisfied. Thus x should converge and x=1.
 
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It can be proved that the sequence of convergents of any simple continued infinite fraction converges.
 
Btw, the continued fraction with all ones is the golden ratio and it converges extremely slowly.
In this respect it is the most irrational of all numbers.
 
Hi guys. I believe I figured this out now. Maybe I should have thought about it more before posting the question. It is possible to show that in this problem we are dealing with a decreasing sequence that's bounded from below. Now I've got a feeling that I don't want to spoil this by giving all the details. I'll leave figuring out the details as a challenge for those who are interested.

I became interested in this representation of 1, because in Wikipedia page on irrationality of Neper's number, I encountered a claim that if a continued fraction doesn't terminate, then it couldn't represent a rational number, and it must be irrational. (Rational numbers always have terminating continued fractions?) Doesn't this equation in my opening post debunk the claim from Wikipedia as false?
 
jostpuur said:
$$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$
This is not a simple continued infinite fraction.

$$ a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{...}}}} $$
All integers in the sequence ## \{a_i\} ##, other than the first, must be positive.
 
anuttarasammyak said:
Say RHS continued fraction is x, we have an equation
x=2−1x
(x−1)2=0
x=1
So we know if x converges, x=1. If it does not converge the equation is not satisfied. Thus x should converge and x=1.
Generalizing 2 to integer n, the equation is
$$x^2-nx+1=0$$
$$x=\frac{n \pm \sqrt{n^2-4}}{2}$$
So we know for x to exist ##|n|\geq 2##

$$x=\frac{n + \sqrt{n^2-4}}{2}$$ for n>2 because x > n - 1/n.
$$x=\frac{n - \sqrt{n^2-4}}{2}$$ for n<-2 because x < n - 1/n.

Is it right ?
 
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