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.