It is my understanding that if a continued fraction is finite, then it is typically written: $$ [a_0; a_1, a_2]_c = a_0 + \frac{1}{a_1 + \frac{1}{a_2}} $$ (where subscript $c$ stands for "current definition") whereas if it is infinite, one writes it as: $$ [a_0; a_1, a_2, ...]_c = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{...}}} $$
To me, however, this ellipsis makes the definitions of finite and infinite continued fractions different. Specifically, with this "definition", I think we have that: $$ [a_0; a_1, a_2, 0, 0, 0, ...]_c = [a_0; a_1, a_2 +1]_c $$ because: $$ \frac{1}{0 + \frac{1}{0 + \frac{1}{...}}} = \frac{1}{\frac{1}{\frac{1}{...}}} = 1 $$
Hence my questions:
- Is this correct? I.e. do we have $[a_0; a_1, a_2, 0, 0, ...]_c = [a_0; a_1, a_2+1]_c$?
- Why not define continued fractions such that $[a_0; a_1, a_2]_p = [a_0; a_1, a_2, 0, 0, ...]_p$ (where $p$ stands for "proposed")? E.g. are there desirable properties that would not be valid with such a definition?
This would be equivalent to $[a_0; a_1, a_2]_p = [a_0; a_1, a_2-1]_c$ with the constraint that the last term in the "current" finite representation always be greater than 1. To expand this definition further, we would have: $$ [a_0; a_1, a_2]_p = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + 1}} $$ to be compared with the definition at the top of this post.
As far as I can see, such a definition would be beneficial. For example, it would eliminate the ambiguity in the finite case: $$ [a_0; a_1, ..., a_{k-1}, 1]_c = [a_0; a_1, ..., a_{k-1}+1]_c $$ which is typically "solved" by imposing $a_{k}\geq 2$. This would be implied with the proposed definition.
It seems to me that this is a "mistake" that has now been accepted, but perhaps I am missing something?