Let $\frac{P_n} {Q_n} and \frac{P_{n+1}} {Q_{n+1}}$ be two consecutive continued fraction convergents for $b$. Then prove that:
$$\left|{\frac{P_n} {Q_n}-b}\right|< \frac{1}{2Q_n^2}$$
or
$$\left|{\frac{P_{n+1}} {Q_{n+1}}-b}\right|< \frac{1}{2Q_{n+1}^2}$$
Thanks in advance!
The consecutive convergents are alternately greater than and less than the value of $b$. Then we have that
$${1\over q_nq_{n+1}}=\left|{p_{n+1}q_n-p_nq_{n+1}\over q_nq_{n+1}}\right|=\left|{p_{n+1}\over q_{n+1}}-{p_n\over q_n}\right|=\left|{p_{n+1}\over q_{n+1}}-b\right|+\left|{p_{n}\over q_{n}}-b\right|$$
The last equality follows because they one is bigger and the other is smaller than $b$, this is where it is essential that they are consecutive convergents.
Now, if the premise is false, then
$${1\over q_nq_{n+1}}\ge {1\over 2q_n^2}+{1\over 2 q_{n+1}^2}$$
which is blatantly impossible.