Quadratic surd continued fraction convergent ratio limit Given a quadratic surd $\sqrt d$ where $d$ is a natural number and not a perfect square. $(c_i)_{i=1}^\infty$ is the sequence of convergents of the continued fraction of $\sqrt d$. Let $s_i:=\frac{c_{i+1}-\sqrt d}{c_i-\sqrt d},\,\forall i\in\mathbf N$. Let $n$ be the period of the continued fraction. Are the followings true?  


*

*$l_r:=\lim_\limits{i\rightarrow\infty}s_{in+r}$ exists for any $r\in\{0,1,\cdots,n-1\}$.

*For $n\le 2$, $l_0=l_1$.

*For $n\ge 3$, there exists at least two distinct $l_r$'s. 

Conjecture 1. is proved below. However, conjectures 2. and 3. still await proofs.
 A: Here is my proof to conjecture #1. The principle idea is inspired and reminded by the example @StevenStadnicki set in his answer, even though it is what I should have realized before that.
Instead of the quadratic surd $\sqrt d$ per se, we look at the equivalent purely periodic continued fraction $\alpha := [\overline{a_1,a_2,\cdots,a_n}]$ of period $n$. Define function
$$L_i(c):=[a_1,a_2,\cdots,a_i, c]=\frac{cp_i+p_{i-1}}{cq_i+q_{i-1}}$$
where $(p_i,q_i)$ is the numerator and denominator of the convergent $c_i$.
Note $c_{(i+1)n+r}=L_n(c_{in+r})$ and $L_n(\alpha)=\alpha$.
$$L_i(u)-L_i(v)=\frac{(u-v)(p_iq_{i-1}-p_{i-1}q_i)}{(uq_i+q_{i-1})(vq_i+q_{i-1})}=\frac{(-1)^i(u-v)}{(uq_i+q_{i-1})(vq_i+q_{i-1})},$$
as
$$p_iq_{i-1}-p_{i-1}q_i=(-1)^i\implies c_i-c_{i-1}=\frac{(-1)^i}{q_iq_{i-1}}. \tag1$$ 
So
$$s_{(i+1)n+r}=\frac{L_n(c_{in+r+1})-L_n(\alpha)}{L_n(c_{in+r})-L_n(\alpha)}=\frac{c_{in+r}q_n+q_{n-1}}{c_{in+r+1}q_n+q_{n-1}}s_{in+r}
=\left(1+\frac{\frac{(-1)^{in+r}}{q_{_{in+r+1}}q_{in+r}}}{c_{in+r+1}+\frac{q_{n-1}}{q_n}}\right)s_{in+r}.\tag2$$
From 
$$q_i=a_iq_{i-1}+q_{i-2}\tag3$$
we have
$$\frac{q_n}{q_{n-1}}=[a_n,a_{n-1},\cdots,a_2].$$
It can be easily seen from Eq. (1) that
$$|c_i-\alpha|<\frac1{q_i^2}.$$
From Eq. (3),
$$q_{in+r}>n^i+r.$$
Actually we have a precise computation of the growth of $(q_k)_{k=1}^\infty$ due to the finiteness of $(a_k)_{k=1}^n$. 
$$\begin{bmatrix}
q_{in+r+1} \\
q_{in+r}
\end{bmatrix}
=\begin{bmatrix}
b_1 & b_2 \\
b_3 & b_4
\end{bmatrix}
\begin{bmatrix}
q_{(i-1)n+r+1} \\
q_{(i-1)n+r}
\end{bmatrix}
=\begin{bmatrix}
b_1 & b_2 \\
b_3 & b_4
\end{bmatrix}^i
\begin{bmatrix}
q_{r+1} \\
q_r
\end{bmatrix}.
$$
$q_{in+r}$ grows precisely exponentially. So from Eq. (3) $r_{in+r}$ converges oscillatorily exponentially as $i\rightarrow\infty$.

Proof to conjecture $2.$:
Define
$$P(x):=\frac{\lfloor\sqrt d\rfloor x+d}{x+\lfloor\sqrt d\rfloor}.$$
Obviously $P(\sqrt d)=\sqrt d$.
By Theorem Plus of Square Root, Continued Fractions and the Orbit of $\frac10$ on $\partial H^2$ by Julian Rose, Krishna Shankar and Justin Thomas. The continued fraction of $\sqrt d$ is of period $2$ iff $P^i(\infty)=c_i$ for all convergents $\{c_i\}_{i=1}^\infty$.
$$\frac{P(c_i)-P(\sqrt d)}{c_i-\sqrt d}=-\frac{\sqrt d-\lfloor\sqrt d\rfloor}{c_i+\lfloor\sqrt d\rfloor}.$$
So
$$l_0=l_1=-\frac{\sqrt d-\lfloor\sqrt d\rfloor}{\sqrt d+\lfloor\sqrt d\rfloor}.$$

Attempt at proving Conjecture 3:
A fraction linear transform $L$ with integer coefficients with fixed points $\pm\sqrt d$ is of the form $L(x)=\frac{ax+cd}{cx+a}$ for some integers $a$ and $c$. Of course $L_n(x)=\frac{(x+a_1)p_n+p_{n-1}}{(x+a_1)q_n+q_{n-1}}$ where $p_i$ and $q_i$ are the numerators and denominators of the convergents, $a_1=\lfloor \sqrt d \rfloor$, and $n$ is the period of the continued fraction, is one such fractional linear transformation. Let $\alpha=\sqrt d+\lfloor \sqrt d \rfloor$ and $\alpha'=\sqrt d-\lfloor \sqrt d \rfloor$ be its conjugate. A linear fractional transformation $f$ preserves the cross ratio, i.e.
$$\frac{(x_3-x_1)(x_4-x_2)}{(x_3-x_2)(x_4-x_1)}=\frac{(f(x_3)-f(x_1))(f(x_4)-f(x_2))}{(f(x_3)-f(x_2))(f(x_4)-f(x_1))}.$$
With $f=L_n$ and $L_n(\alpha)=\alpha$, $L_n(\alpha')=\alpha'$ we have 
$$\frac{(c_{r+1}-\alpha)(c_r-
\alpha')}{(c_r-
\alpha)(c_{r+1}-\alpha')}=\frac{(L_n^k(c_{r+1})-\alpha)(L_n^k(c_r)-\alpha')}{(L_n^k(c_r)-\alpha)(L_n^k(c_{r+1})-\alpha')},\ \forall k\in\mathbf N$$
From the proof of Conjecture #1 and $\lim_{k\to\infty}L_n^k(c_r)=\alpha$, we know the right hand side converges to $l_r$ as $k\to\infty$. So we have an explicit expression for
$$l_r = \frac{(c_{r+1}-\alpha)(c_r-
\alpha')}{(c_r-
\alpha)(c_{r+1}-\alpha')}.$$
