I haven't taken any number theory courses but out of curiosity I learned about [Dirichlet's approximation theorem](https://en.wikipedia.org/wiki/Dirichlet's_approximation_theorem). Afterwards, it occurred to me to define the following function $f$ using 'optimal' diophantine approximants:


$\forall \alpha \in \mathbb{R}\_+ \setminus \mathbb{Q}$, let $\{\alpha\_q\}\_{q=1}^\infty \subset \mathbb{Q}$ denote the
optimal diophantine approximants of $\alpha$:


\begin{equation}
\forall q \in \mathbb{N}, |\alpha\_q -\alpha| = \min\_{n \in \mathbb{N}}|\frac{n}{q}-\alpha|
\end{equation}


Using the language of functions rather than sequences we may define:


\begin{equation}
\begin{cases}
f: \mathbb{R}\_+ \setminus \mathbb{Q} \rightarrow \mathbb{R}\_+ \\
f(\alpha) = \sum\_{q=1}^{\infty} |\alpha\_q-\alpha|
\end{cases}
\end{equation}


My question is whether $\forall x \in \mathbb{R}\_+ \setminus \mathbb{Q},f(x)< \infty $


If so, is there an elementary proof?



Beyond Dirichlet's approximation theorem
Since $\alpha$ is irrational, the [equidistribution theorem](https://en.wikipedia.org/wiki/Equidistribution_theorem) says that the fractional parts $\{q\alpha\}$ are uniformly distributed in $[0,1)$. In particular, a positive proportion of $q$ satisfy $\{q\alpha\} \in [\frac13,\frac12]$. For these $n$, we have $|\alpha\_q-\alpha| = \frac{\{q\alpha\}}q \ge \frac1{3q}$, and so $f(\alpha)$ is at least as large as the sum of $\frac1{3q}$ over a positive proportion of $q$, which diverges. (Note that the sum also diverges when $\alpha$ is rational.)


