That's correct. The partial quotients of a continued fraction expansion of the square root of a non square integer $n$, written $\sqrt{n}=[a_0;\overline{a_1, a_2, ..., a_{k-1},2a_0}]$, with primitive period $k_p$, have the property $a_i\leqslant a_0$ for $k_p\nmid i$. For non primitive period, $a_i\leqslant 2a_0$, when $i\geqslant 0$.
Proof.
Let the complete quotients of the expansion of the continued fraction of $\sqrt{n}$ be denoted $r_h=\dfrac{\sqrt{n}+A_{h-1}}{\alpha_{h-1}}$, with $r_0=\sqrt{n}$.
Using the following properties ($h\geqslant 0$, $j\geqslant -1$):
$\;\;\;\;\;\;\;\;(1)\;\;\;\;\;\;\;\;A_j\leqslant a_0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$(Perron p.76 formula 7)
$\;\;\;\;\;\;\;\;(2)\;\;\;\;\;\;\;\;A_{h-1}+A_h=a_h\alpha_{h-1}\;\;\;\;\;$ (Perron p.83 formula 4)
$\;\;\;\;\;\;\;\;(3)\;\;\;\;\;\;\;\;\alpha_{i-1}\geqslant 2 \text{ for } 0<i<k_p\;\;$ (Perron p.93 formula 5)
(Note:above indices are shifted for $A_h$ and $\alpha_h$ compared to Perron's notations)
(2) and (1) gives $a_h\leqslant \dfrac{2a_0}{\alpha_{h-1}}$, and using (3) we get $a_i\leqslant a_0$ for $0<i<k_p$ and by periodicity, for all $a_i$ with $k_p\nmid i$. And since all $\alpha_{h-1}$ for $h\geqslant0$ are positive integers, we have $a_h\leqslant 2a_0$ for $h\geqslant0$, the period being primitive or not.
Die Lehre von den Kettenbrüchen (Perron)
Best Answer
Note: Some people start their continued fractions with index $1$, we adopt this convention, does not change the content.
Consider the fraction
$$[a_1, a_2, \ldots, a_n, x] = \frac{p x + p'}{q x + q'}$$
where $\frac{p}{q} = [a_1, \ldots, a_n]$, and $\frac{p'}{q'} = [a_1, \ldots, a_{n-1}]$.
We see that including $x$ in a continued fraction with entries $[a_1, \ldots, a_n]$ applies to $x$ a certain Möbius transform. It is given by the matrix $\begin{pmatrix} p & p' \\ q & q'\end{pmatrix}$. But one sees easily that it is in fact the composition of Mobius transforms like this $t \mapsto a + \frac{1}{t}$, which is given by the matrix $\begin{pmatrix} a &1 \\1 & 0 \end{pmatrix}$. Therefore, we have
\begin{eqnarray} \begin{pmatrix} p & p' \\ q & q'\end{pmatrix} = \begin{pmatrix} a_1 & 1 \\ 1 & 0\end{pmatrix}\cdot \begin{pmatrix} a_2 & 1 \\ 1 & 0\end{pmatrix}\cdot \cdots \cdot \begin{pmatrix} a_n & 1 \\ 1 & 0\end{pmatrix} \end{eqnarray}
Now, your number $\theta$ is the positive root of the equation
$$x = \frac{ p x + p'}{q x + q'}$$
How to get the other number?
In the equation above with products of matrices, take the transpose on both sides. Notice that on RHS we have symmetric matrices. Therefore
\begin{eqnarray} \begin{pmatrix} p & q \\ p & q'\end{pmatrix} = \begin{pmatrix} a_n & 1 \\ 1 & 0\end{pmatrix}\cdot \begin{pmatrix} a_{n-1} & 1 \\ 1 & 0\end{pmatrix}\cdot \cdots \cdot \begin{pmatrix} a_1 & 1 \\ 1 & 0\end{pmatrix} \end{eqnarray}
We conclude that $\psi = [\overline{a_n, a_{n-1}, \ldots, a_1}]$ is the positive root of the equation
$$x = \frac{p x + q}{p' x + q'}$$
Therefore
$$\theta = \frac{(p-q') + \sqrt{(p+q')^2 - 4 ( p q' - p' q)}}{2 q}$$
while $$\psi = \frac{(p-q') + \sqrt{(p+q')^2 - 4 ( p q' - p' q)}}{2 p'}$$
We notice that $\psi = \frac{q}{p'} \cdot \theta\ \ $. Also check that $$\theta \cdot \bar {\theta} = - \frac{q'}{p}$$
Therefore we get
$$\psi = - \frac{\theta}{N( \theta)}$$
quite an inversion.