Question

Is there any simpler way of writing `1/(x+1/((x+1)+1/((x+2)+1/((x+3)+ddots`, where `x=RR`?

For example, `1/(1+1/(2+1/(3+1/(4+ddots`

Answer 1

In terms of notation, sure. If you wanted to denote an infinite continued fraction, i.e.

`x + 1/(x+1 + 1/(x+2 + 1/(x + 3 + 1/(x + 4 + ddots)))`,

you would write

`[x; x + 1, x + 2, x + 3, . . . ]`

For example, if you represent `425/27`, a finite continued fraction, you could write

`15 + 20/27`,

or by multiplying through by `(1/20)/(1/20)`, and then `(1/7)/(1/7)` on the inside, etc., until all nested numerators are `1`:

`(425)/27 = color(red)(15) + 1/(color(red)(1) + 1/(color(red)(2) + 1/(color(red)(1) + 1/color(red)(6)))`,

or

`[15; 1, 2, 1, 6]`

So, for your example, to write

`1/(1 + 1/(2 + 1/(3 + 1/(4 + ddots))))`,

you could write

`[0; 1, 2, 3, 4, . . . ]`

Answer 2

See below.

Explanation:

Calling `f(x) = 1/(x+1/((x+1)+1/((x+2)+1/((x+3)+ddots` we can build the recurrence equation

`f(x) = 1/(x + f(x+1))` which gives a nonlinear functional equation as

`f(x)(f(x+1)+x)-1 = 0`