To which $\alpha\in\mathbb{R}$ can $\frac{p}{q}$ be a convergent of its (alpha's) continued fractions? I read that for any to consecutive convergents of a number $\alpha$, at least one of them must be distance at most $\frac{1}{2q^2}$ from $\alpha$. I don't see how this helps me into determining in what interval $\alpha$ can lie when you know $\frac{p}{q}$ is a convergent of its continued fractions. I would say that $\alpha$ is in $\left(\frac{p}{q}-\frac{1}{2q^2},\frac{p}{q}+\frac{1}{2q^2}\right)$, but I'm not sure and I don't see whether I can used the mentioned theorem for this. Any suggestions?
To state the problem a bit more clear: given $\frac{p}{q}=[a_0,\ldots,a_n]$, to what $[a_0,\ldots,a_n,a_{n+1},\ldots]=\alpha\in\mathbb{R}$ can this fraction be a convergent of its continued fraction? I would say that these $\alpha$ lie in some interval.
 A: EDIT: This answer is incorrect, but since it has been accepted, I can't delete it. See the answer by @dennis.
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I think I understand the question now, and the answer is this:
Let $p_n/q_n=[a_0;a_1,\dots,a_n]$, and let $p_{n-1}/q_{n-1}=[a_0;a_1,\dots,a_{n-1}]$. Then $p_n/q_n$ is a convergent to the continued fraction of $\alpha$ if and only if $\alpha$ is between $p_n/q_n$ and ${p_n\over q_n}+{(-1)^n\over q_n(q_n+q_{n-1})}$. Note that $\alpha$ must be bigger than $p_n/q_n$ if $n$ is even, and smaller if $n$ is odd. Note also that $q_n(q_n+q_{n-1})$ is between $q_n^2$ and $2q_n^2$, but exactly where in between depends on $q_{n-1}$.
A: An equivalent but more compact statement is given is Theorem 2 of
https://hal.archives-ouvertes.fr/hal-02272389
(paper written in french).
Let $p$ and $q \ge 2$ be two relatively prime integers. Call $(u,v)$ the unique solution of the Bezout equation $up-vq=1$, with $0 \le u \le q-1$
(then $u \ge 1$). Let $\theta \in \mathbb{R}$. Then:
$\bullet$ $p/q$ is a convergent or a semi-convergent of $\theta$ if an only if $\frac{v}{u} < \theta < \frac{p-v}{q-u}$.
$\bullet$ $p/q$ is a convergent of $\theta$ if an only if $\frac{p+v}{q+u} < \theta < \frac{2p-v}{2q-u}$.
For example, if $p=3$ and $q=10$, then $u=7$ and $v=2$. Hence:
$\bullet$ $3/10$ is a convergent or a semi-convergent of $\theta$ if an only if $2/7 < \theta < 1/3$.
$\bullet$ $3/10$ is a convergent of $\theta$ if an only if $5/17 < \theta < 4/13$.
Remark: when $p>0$, the fractions $v/u$ and $(p-v)/(q-u)$ are the two fractions providing the fraction $p/q$ in the Stern-Brocot tree. And the fractions $(p+v)/(q+u)$ and $(2p-v)/(2q-u)$ are the two chidren of the fraction $p/q$ in the Stern-Brocot tree.
https://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree
