Consider a simple continued fraction

$a_{0}\in \mathbb{Z} , a_{n}\in \mathbb{Z}_{\geq 0}, \quad \xi=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\ddots}}}=\left[a_0, a_1, \dotsc\right].$

Define \begin{aligned} & h_{-2}=0 \quad h_{-1}=1 \quad h_n=a_n h_{n-1}+h_{n-2}
& k_{-2}=1 \quad k_{-1}=0 \quad k_n=a_n k_{n-1}+k_{n-2}.
\end{aligned}

Then $\quad a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\dotsb +\cfrac{1}{a_n}}}}=\left[a_0, a_1, \dotsc, a_n\right] =\frac{h_n}{k_n}$ and $\frac{h_n}{k_n} \rightarrow \xi$.

Then the following hold:

$\sum_{n=-1}^{\infty}\left|k_n \xi-h_n\right| \cdot a_{n+1}=\xi+1$ for an irrational $\xi$

$\sum_{n=-1}^N\left|k_n \xi-h_n\right| a_{n+1}=\xi+1-\frac{1}{p}$ for a rational $\xi=q/p=[a_0,\dots,a_{N+1}]$

$\sum_{n=-1}\limits^{\infty}\left|k_n \xi-h_n\right|^2 \cdot a_{n+1}=\xi$ if $\xi$ is irrational and
$\sum_{n=-1}\limits^{N}\left|k_n \xi-h_n\right|^2 \cdot a_{n+1}=\xi$ for $\xi=[a_0,\dots,a_{N+1}]$.

I know at least two elementary proofs of these formulae. I believe that these formulae were discovered before. Have you seen something like that?

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