I was on the same article mentioned on here Continued fraction rate of convergence.
What I did not understand was this identity used to prove the right hand side of the inequality.
$$ \left| \alpha -\frac{p_n}{q_n} \right|< \frac{1}{q_n q_{n+1}} $$
For the RHS, there is a theorem that says
$$ \frac{p_{n-1}}{q_{n-1}} - \frac{p_n}{q_n} = \frac{(-1)^n}{q_n q_{n-1}} $$
So intuitively, I have that $\frac{1}{q_n q_{n+1}} = \left| \frac{p_{n}}{q_{n}} - \frac{p_{n+1}}{q_{n+1}} \right| \implies \left| \alpha -\frac{p_n}{q_n} \right| < \left| \frac{p_{n}}{q_{n}} - \frac{p_{n+1}}{q_{n+1}} \right|$
Assuming that the identity $ \left| \alpha -\frac{p_n}{q_n} \right|< \frac{1}{q_n q_{n+1}} = \left| \frac{p_{n}}{q_{n}} - \frac{p_{n+1}}{q_{n+1}} \right|$ holds true, it doesn't make sense to me. It is telling me that the difference from the $n$-th approximation to $\alpha$ is less than the difference from the $n$-th approximation to the $(n+1)$-th approximation.
If it is true, how can I dervive it?