= Solution
Set
$$
p_{-2}=0, p_{-1}=1,\qquad q_{-2}=1, q_{-1}=0,
$$
and recursively
$$
p_n=a_np_{n-1}+p_{n-2},\qquad q_n=a_nq_{n-1}+q_{n-2}.
$$
Then $p_n/q_n=[a_0,ldots,a_n]$. The <determinant> identity
$$
p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1}
$$
follows by induction. Writing the remaining complete quotient as $\theta_{n+1}>1$ gives
$$
\theta=\frac{p_n\theta_{n+1}+p_{n-1}}
{q_n\theta_{n+1}+q_{n-1}},
$$
so $\theta$ lies strictly between consecutive convergents and their order alternates. For odd $n$,
$$
\frac{p_{n-1}}{q_{n-1}}<\theta<\frac{p_n}{q_n}.
$$
Moreover $|\theta-p_n/q_n|<1/(q_nq_{n+1})$, and $q_n\to\infty$, proving convergence.
The continued-fraction algorithm gives
$$
\boxed{\sqrt7=[2;\overline{1,1,1,4}].}
$$
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