# Continued fraction question

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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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. – André Nicolas Jun 19 '12 at 1:21

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 $1+\frac1x = 1+\frac1{1+}\frac1{1+}\cdots = x$.
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.