Lately I was reading a bit about continued fractions and came up with a question that I couldn't find an answer for.
Here are the definitions I will use in the question:
Let $x \in \mathbb{R}$. A fraction $\frac{p}{q}$ (assume $q > 0$) is said to be a rational best approximation of $x$ if $$\left| x – \frac{p}{q}\right| \leq \left|x – \frac{p'}{q'}\right|$$ for all $p', q' \in \mathbb{Z}, 1 \leq q' \leq q$.
Then $\frac{p}{q}$ is called a good approximation of $x$ if
$$ \left|x – \frac{p}{q}\right| < \frac{1}{q^2}. $$
Now I know that every convergent of the continued fraction for $x$ is both a best approximation and a good approximation.
On the other hand: Not every best approximation for $x$ is given through a convergent of its continued fraction (take e.g. $13/4$, which is not a convergent of $\pi$ but a rational best approximation).
My question is: Is every good approximation given through a convergent? By checking a few examples of best rational approximations which are no convergents I got the feeling that this could be true, but I did not find a definite answer on this.
Best Answer
The first few convergents for $\sqrt{3}$ are $1,2,5/3,7/4,19/11.$ In particular none have denominator $7,$ but $\sqrt{3}-12/7 \approx 0.0177$ while $1/49 \approx 0.0204.$ So in this example we have a good approximation which is not a convergent.
Added: If one defines a "very good" approximation of an irrational $x$ as one for which $|x-p/q|<1/(2q^2),$ then it is known that any very good approximation to an irrational must be one of the convergents to it. The above example of the good approximation $12/7$ is not close by a margin of $1/2\cdot 49$ to $\sqrt{3},$ as would be expected by this known result.