For any real number $r$, the convergents to the continued fraction expansion of $r$ satisfy Dirichlet's approximation inequality of $|r – \frac{p}{q}| < \frac{1}{q^2}$.
Does this go the other way? That is, if given some rational number $p/q$, we have that $|r – \frac{p}{q}| < \frac{1}{q^2}$, does that necessarily mean that it is a convergent of the continued fraction expansion of $r$?
It does seem, from this page, that we have a similar result that the convergents, and only the convergents, minimize $|r – \frac{p}{q}| \cdot q$ among all smaller rationals, so we would need to show that whenever this happens, we also have $|r – \frac{p}{q}| \cdot q < \frac{1}{q}$.
EDIT: Legendre has also proven that if $|r – \frac{p}{q}| < \frac{1}{2q^2}$ then $p/q$ is a convergent of $r$. So the main question is what happens if for whatever $p/q$ we have that $\frac{1}{2q} < |r – \frac{p}{q}| < \frac{1}{q}$: can we determine that $p/q$ is a convergent, or a semiconvergent, or anything at all?
Best Answer
Three results summarize how good continued fractions are at approximating. Here, suppose that $\frac{p_0}{q_0}, \frac{p_1}{q_1}, \frac{p_2}{q_2}, \dots$ are the continued fraction approximants to $r \notin Q$.
However, there are other approximations that are good enough for Dirichlet, but not good enough for Legendre. As pointed out in the comments by @halbaroth, there is a fourth result that characterizes such approximations:
The intuition here is that the continued fraction approximants satisfy the recursion $\frac{p_n}{q_n} = \frac{a_n p_{n-1} + p_{n-2}}{a_n q_{n-1} + q_{n-2}}$. So we often get some extra approximations along the way by replacing $a_n$ with some $k \in \{1, 2, \dots, a_n -1\}$. These are all okay approximations, but the only ones that meet Fatou's standards are $k=1$ and $k = a_n - 1$: the ones that are closest to $\frac{p_{n-1}}{q_{n-1}}$ or $\frac{p_n}{q_n}$.
For example, the first few continued fraction approximants to $\sqrt{2}$ are $$1,\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\frac{99}{70},\dots$$ The first few fractions $\frac pq$ with $|\sqrt2 - \frac pq| < \frac1{q^2}$ are $$1, \frac32, \color{red}{\frac43}, \frac75, \color{red}{\frac{10}7}, \frac{17}{12}, \color{red}{\frac{24}{17}}, \frac{41}{29}, \color{red}{\frac{58}{41}}, \frac{99}{70}, \dots$$ The extra fractions appearing in the gaps are the ones Fatou suggests: for example, $\frac{58}{41} = \frac{41 + 17}{29+12} = \frac{99-41}{70-29}$.
In the case of $\sqrt2$, the error $|\sqrt2 - \frac pq|$ turns out to be approximately $\frac1{2\sqrt2 q^2}$ for the fractions in black and approximately $\frac1{\sqrt2 q^2}$ for the fractions in red, so all of these are good enough.
This does not have to happen; in the case of $\pi$, for example, in between $\frac{22}{7}$ and $\frac{333}{106}$ we have a sequence of approximations $\frac{25}{8}, \dots, \frac{311}{99}$. Fatou promises us only the first and last of these have a chance of being good, but actually, $|\pi - \frac{25}{8}| \approx \frac{1}{0.94 \cdot 8^2}$ and $|\pi - \frac{311}{99}| \approx \frac1{0.57 \cdot 99^2}$: neither is good enough. Fatou does not promise that any of these approximations will satisfy the inequality $|r - \frac pq| < \frac1{q^2}$ - only that nothing else will.
Additionally, the approximations $\frac{p_{n+1}+p_n}{q_{n+1}+q_n}$ and $\frac{p_{n+2}-p_{n+1}}{q_{n+2}-q_{n+1}}$ can, in some cases, get arbitrarily close to the constant of Legendre's theorem. For this, consider $x = 50 + 5\sqrt{102}$: this has continued fraction expansion $[100;2,100,2,100,2,\dots]$. (I hope it is clear how I got this example.) Two consecutive continued fraction approximants to $x$ are $\frac{40601}{404}$ and $\frac{4080300}{40601}$. The first of these is very good: $404^2|x - \frac{40601}{404}| \approx \frac{1}{101}$. The second of these is still decent: $40601^2|x-\frac{4080300}{40601}| \approx \frac1{2.02}$. Well, if we add these together, the result $\frac{4120901}{41005}$ is nearly as good: $41005^2|x - \frac{4120901}{41005}| \approx \frac1{1.98}$.