Inversion of infinite continued fractions

[denijane]

Hello to everyone!
I need to invert the following infinite continued fraction:
0=\beta_{0}-\frac{\alpha_{0}\gamma_{1}}{\beta_{1}-}\frac{\alpha_{1}\gamma_{2}}{\beta_{2}-}\ldots, n=1..\infty<br />
to something starting with the \beta_{n} term and going down to n=1 (where n is the term the fractions will be cut to).

I know there is the following identity for finite continued fractions:
<br /> \frac{[a_{0},\ldots, a_{n}]}{[a_{0},\ldots, a_{n-1}]}=\frac{[a_{n},\ldots, a_{0}]}{[a_{n-1},\ldots, a_{0}]}<br /> http://mathworld.wolfram.com/ContinuedFraction.html"

But I'm not sure if it remains true for truncated infinite continued fractions.
Any help or resources will be appreciated.

 

[KevB]

If
x = [a_{0} ; a_{1}, a_{2},...]

1/x = [ 0 ; a_{0}, a_{1}, a_{2}, ... ] \ when \ a_{0} \neq 0
\ \ \ \ \ \ = [ a_{1}; a_{2}, a_{3},... ] \ when \ a_{0} = 0