Sum of Continued Fractions

Question

Let $x$ be a positive integer. Consider the following sum (maybe there is a better notation with continued fractions, but I am not aware of it):

$\frac{1}{x} + \frac{1}{x- \frac{1}{x}} + \frac{1}{x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}}} +\frac{1}{x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}} - \frac{1}{x- \frac{1}{x} - \frac{1}{x-\frac{1}{x}}}} + \dots $

My question is: How many summands do I need such that the sum is at least $x$? $x^2$ is a trivial upper bound here, because every summand is $\geq \frac{1}{x}$. Is there a way to get a better bound here?

Answer 1

The number of steps needed, $s(x)$, is the sequence A140949 in OEIS. Not sure about the background. Also there is not much information there. Perhaps the auther Gil Broussard can tell you more (assuming that he is not you).

However it does look like there is some pattern going on here. The value of $s(x)$ is very close to $x^2 / 2$. In fact, the sequence $(2s(x) - x^2)_{x \geq 0}$ looks like this:

$$0, 1, 2, 3, 4, 3, 4, 3, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7$$

This looks quite interesting and is probably a logarithmic growth. Thus $x^2 / 2$ is a fairly good approximation of $s(x)$. Not sure about the theory behind.

Also not sure what happens for non-integer real numbers $x$.

I will update this answer if I am motivated enough to look deeper into the problem. That is, if nobody gives a better reference.

It would also help if you could tell us your motivation, e.g. where does this problem come from.

Answer 2

If $\{a_i\}$ is the sequence of terms, and $s_n$ is the sum of first $n$ terms, then we get $$a_{n+1} = \frac{1}{x-s_{n}}$$ for $n\geq1$ and $a_1= \frac{1}{x}$. Rearranging this, $s_n = x-\frac{1}{a_{n+1}}$. Each term is $\geq 1/x$, so their reciprocals are $\leq x$.