Convergence
In some sense, the convergents are the best possible approximations for a given nonnegative real:
Theorem: Let \(p/q\) be a convergent for the nonnegative real \(a\). Then if \(p'/q'\) is a closer rational approximation, then \(q' \gt q\).
Proof: Recall successive convergents get closer to \(x\) and alternatively overshoot and undershoot \(x\). Also recall \(p_k q_{k1}  q_k p_{k1} = (1)^k\). The result follows from the next lemma.∎
Lemma: Suppose
and \(a b'  a' b = 1\). Then \(d \gt b\) and \(d \gt b'\).
Proof: Manipulating the inequalities gives
As \(c b  a d \gt 0\), we find \(b' \lt d\). The proof for \(b\) is similar.∎
We can say more about any rational approximation to \(a\). Suppose \(k\) is odd. The following generalizes to even \(k\) by flipping signs.
Since \(p_{k+1}, q_{k+1}\) are coprime, the solutions of \(p_{k+1} x  q_{k+1} y = 1\) are given by \(x = q_k + t q_{k+1}, y = p_k + t p_{k+1}\) for all integers \(t\). Likewise, we can construct a sequence of approximations, the intermediate convergents,
This sequence strictly increases, as do all the numerators and denominators, and gets closer and closer to \(p_{k+1}/q_{k+1}\).
Given a rational approximation \(p'/q'\) to \(a\), we have three cases:

\(p'/q'\) is one of the intermediate convergents.

\(p'/q'\) lies between two of the intermediate convergents, hence by the lemma \(q'\) is greater than the denominator of the intermediate convergents on either side.

\(p'/q'\) lies between the last intermediate convergent and \(a\). By the theorem \(q' \gt q_{k+2}\).
We establish handy rules of thumb for the accuracy of a convergent. Let \(x_k = [a_k; a_{k+1},...]\). Recall
thus
Furthermore \(x_{k+1}q_k + q_{k1} \ge a_{k+1} q_k + q_{k1} = q_{k+1} \ge q_k \), with equality only when \(x\) is a rational terminating with \(a_{k+1}\). We also have
In summary,
Theorem: Let \(p_k/q_k\) be the convergents of a nonnegative real \(a\). Then
Now for a sort of converse:
Theorem: If \(\left \frac{r}{s}  a \right \lt \frac{1}{2s^2}\) where \(r,s\) are coprime then \(\frac{r}{s}\) is a convergent of \(a\).
Proof: Let the expansion of \(r/s\) be \([a_1; a_2,...,a_k]\) where \(k\) is odd. (Recall a rational has two possible expansions, one exactly one term longer than the other.) Define \(x_{k+1}\) by the real that satisfies \(a = [a_1;a_2,...,a_k, x_{k+1}]\).
We will show \(x_{k+1} \ge 1\), because this implies we could expand \(x_{k+1}\) into a continued fraction with a nonzero first term and obtain a continued fraction expansion for \(a\), proving the theorem.
Rewriting
gives
As \(k\) is odd, we have \(x  \frac{p_k}{q_k} = \epsilon\) for some \(0 \lt \epsilon \lt \frac{1}{2q_k^2}\), thus
We need only show \(1  \epsilon q_k q_{k1} \ge \epsilon q_k^2\):
At last we plug a hole in our proof that rationals have exactly two finite continued fraction expansions. Suppose the rational \(p/q\) has an infinite continued fraction expansion. This would imply
as \(n \rightarrow \infty\), a contradiction.
See also Farey sequences and SternBrocot trees.