<s>[INST] Sum of Continued Fractions 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?
 [/INST] 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 $\leq x$.
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