Show that an irrational number $\xi= [a_0;a_1,\cdots]$ has bounded partial quotients $a_k$ if and only if there exists $\delta >0$ such that $$\left| \xi - \frac{p}{q} \right| > \frac{\delta}{q^2}$$ for all rational $\frac{p}{q}$.
Since $\xi$ is irrational, $\xi$ can be expanded into an infinite regular continued fraction, which is unique. Moreover, using best approximations, if $\xi$ is irrational then $$\left| \xi - \frac{p_k}{q_k} \right| < \frac{1}{q_k^2}$$ since $q_{k+1} < q_k(a_{k+1}+1)+q_{k-1} = q_{k+1}+q_k$ and $\{q_k\}$ is an increasing sequence. We can find the convergents through computation, but I wanted to know how to formally show the problem at hand.
Take $\delta < 1/2$. If $$\left| \xi - \frac{p}{q} \right| < \frac{\delta}{q^2}$$ then $p/q = p_k/q_k$ is a convergent of the continued fraction expansion of $\xi$ and $$\frac{1}{q_k(q_k+q_{k+1})} < \left| \xi - \frac{p_k}{q_k} \right| < \frac{1}{q_kq_{k+1}}.$$ But $a_{k+1}q<q^2<(a_{k+1}+1)q^2$ since $q_{k+1} = a_{k+1}q_k + q_{k-1}.$ If $a_k < M$ for some $M$ then $$\left| \xi - \frac{p}{q} \right| > \frac{\delta}{q^2}$$ for $\delta = \frac{1}{M+1}.$ This proves one direction.
Conversely, suppose $$\left| \xi - \frac{p}{q} \right| > \frac{\delta}{q^2}.$$ Then $a_k < 1/{\delta},$ so that $a_k$ is bounded.