The condition being that there exists a sequence of rational numbers $\frac{p_i}{q_i}$ such that $$\left|x-\frac{p_i}{q_i}\right|<\frac{1}{3q_i^2}$$
And also that $\frac{p_i}{q_i}\ne\frac{p_j}{q_j}$ for all $i\ne{j}$, $p_1<100$ and $0<p_{i+1}<100p_i$.
From a claim in my homework assignment, I can deduce all $\frac{p_i}{q_i}$ appear in $x$'s sequence of convergents. However I don't know how to proceed. Any advice would be appreciated.
EDIT: I have made some progress by being able to show that $a_k<100$:
$$a_{k}=\frac{p_{k}-p_{k-2}}{p_{k-1}}<\frac{100p_{k-1}-p_{k-2}}{p_{k-1}}=100-\frac{p_{k-2}}{p_{k-1}}<100$$
Now for any $p,q$, we can take $q_k>q$ and since among all rational approximations, with denominators smaller than a convergent's, we have
$$\left|x-\frac{p_{k}}{q_{k}}\right|<\left|x-\frac{p}{q}\right|$$ And then since for all $k>0$:
$$\left|x-\frac{p_k}{q_k}\right|>\frac{1}{\left(a_{k+1}+2\right)q_k^2}$$
we get
$$\frac{1}{200q_{k}^{2}}<\frac{1}{\left(a_{k+1}+2\right)q_{k}^{2}}<\left|x-\frac{p_{k}}{q_{k}}\right|<\left|x-\frac{p}{q}\right|$$
But that doesn't quite get me where I need to be. Is this a good direction? What am I missing?
EDIT: I've found a solution. Thanks for everybody's input.
Edited with my solution. Slowly but surely I got there. :)