Reciprocal of a continued fraction I have to prove the following:
Let $\alpha=[a_0;a_1,a_2,...,a_n]$ and $\alpha>0$, then $\dfrac1{\alpha}=[0;a_0,a_1,...,a_n]$
I started with
$$\alpha=[a_0;a_1,a_2,...,a_n]=a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{a_3+\cfrac1{a_4+\cdots}}}}$$
and
$$\frac1{\alpha}=\frac1{[a_0;a_1,a_2,...,a_n]}=\cfrac1{a_0+\cfrac1{a_1+\cfrac1{a_2+\cfrac1{a_3+\cfrac1{a_4+\cdots}}}}}$$
But now I don't know how to go on. In someway I have to show, that $a_0$ is replaced by $0$, $a_1$ by $a_0$ and so on.
Any help is appreciated.

Coming in late:  there's a similar approach that will let you take the reciprocal of nonsimple continued fractions as well.

*

*change the denominator sequence from $[b_0;a_0,a_1,a_2...]$ to  $[0; b_0,a_0,a_1,a_2...]$

*change the numerator sequence from $[c_1,c_2,c_3,...]$  to $[1,c_1,c_2,c_3...]$
