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I have been given an continued fraction for a number x:

$$x = 1+\frac{1}{1+}\frac{1}{1+}\frac{1}{1+}\cdots$$

How can I show that $x = 1 + \frac{1}{x}$? I played around some with the first few convergents of this continued fraction, but I don't get close.

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  • $\begingroup$ At the formula level, just write down the continued fraction, long-hand, as a long fraction. Then if we let the "value" be $x$, then $x-1$ is $1$ divided by $x$. A formal proof requires some theory. $\endgroup$ Jun 19 '12 at 1:21
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Just look at it. OK, if you want something more proofy-looking: if $x_n$ is the $n$'th convergent, then $x_{n+1} = 1 + 1/x_n$. Take limits.

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Doesn't this immediately follow from the definition of the $n+\frac1{a+}\frac1{b+}\cdots$ notation you are using? Specifically, I thought that $\frac1{a+}Z\ldots$ was defined to be exactly the same as $\frac1{a+Z\ldots}$.

Then if $x=1+\frac1{1+}\frac1{1+}\cdots$ then $\frac1x = \frac1{1+}\frac1{1+}\cdots $ and $1+\frac1x = 1+\frac1{1+}\frac1{1+}\cdots = x$.

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  • $\begingroup$ I feel dumb -.- $\endgroup$
    – Nga
    Jun 19 '12 at 1:21
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    $\begingroup$ Please do not feel dumb. Sometimes a notation is good precisely because it hides this sort of detail so that you can concentrate on more important matters. But until you get used to it, the notation can be paradoxically confusing precisely because it hides the detail. $\endgroup$
    – MJD
    Jun 19 '12 at 1:22

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