Limit of continued fraction Let $(c_n)$ be a sequence of positive numbers converging to $1$ as $n\to\infty$. That is, $c_n > 0$ for every $n$ and $\lim\limits_{n\to\infty}c_n=1$. Let $g_1=-c_1$ and define, for $n \geq 1$,$$g_{n+1}=\frac{-1}{c_n-g_n}.$$
Question: Does $\lim\limits_{n \to \infty} g_n$ exist?
Notice that in case $g_n$ converges to, say $g_\infty$, the limit is given by the negative root of the quadratic equation determined by the equation $g_\infty = \dfrac{-1}{1-g_{\infty}}$.
Notice also that each $g_{n+1}$ can be represented as the following finite continued fraction 
$$g_{n+1}=-\cfrac{1}{c_n+\cfrac{1}{c_{n-1}+\cfrac{1}{\ddots+\cfrac{1}{c_1-g_1}}}},$$
the shorthand for which is sometimes given by the notation $g_{n+1}=[c_n,c_{n-1},\cdots, c_1-g_1]$.
The question is: Does $\lim\limits_{n \to \infty} g_n$ exist?
 A: The way I would do this is as follows: first, assume $c_n=1$, for all $n$; then it's easy to show that $g_{2n}$ decreases, $g_{2n+1}$ increases, $-\frac{1}{2} \geq g_{2k} \geq g_{\infty} \geq g_{2m+1} \geq -1, |g_{2n}-g_{2n+1}| \to 0$ by standard manipulations, so $g_n \to g_{\infty}$ in this case.
Then wlog assume $|c_n-1| \leq \frac{1}{100}$ say for all $n \geq 1$ or start from a place where that happens and renumber if you wish, so in particular $c_n \geq \frac{99}{100}$, call $h_n$ the sequence from step above (where $c_n=1$) and note that $g_n, h_n <0, h_n \leq -\frac{1}{2}, h_n \to g_{\infty}$ from the above. Note also that $|g_{n+1}| \leq \frac{1}{c_n} < 2$ from our assumptions.
Then using $c_n-g_{n} \geq \frac{99}{100}, 1-h_n \geq \frac{3}{2}$ and $|g_{n+1}-h_{n+1}| = |\frac{(c_n-1)+(h_n-g_n)}{((c_n-g_n)(1-h_n)}| \leq  \frac{|c_n-1|+|h_n-g_n|}{\frac{3}{2}\frac{99}{100}}$, it follows that $|g_{n+1}-h_{n+1}| \leq \frac{3}{4}(|c_n-1|+|h_n-g_n|)$. Let $d_n=|h_n-g_n| \leq |h_n|+|g_n| \leq 3, q_n=|c_n-1|, d_n, q_n \geq 0, q_n \to 0 $ and $d_{n+1} \leq \frac{3}{4}(q_n+ d_n)$; then a standard argument presented below shows $d_n \to 0$ so $g_n-h_n \to 0$, so $g_n \to g_{\infty}$. 
Noting that $d_{n+k+1} \leq (\frac{3}{4})^kd_n+(\sup_{p\geq n}q_p)\Sigma{(\frac{3} {4})^m} \leq 3(\frac{3}{4})^k + 4(\sup_{p\geq n}q_p)$, for any $\epsilon >0$, we pick first $n_0$, s.t. $(\sup_{p\geq n_0}q_p) < \frac{\epsilon}{8}$ and then $k_0$ st. $3(\frac{3}{4})^{k_0} < \frac{\epsilon}{2}$ and then $d_{m} < \epsilon, m>n_0+k_0$, so $d_n \to 0$ indeed and we are done.
A: 
Lemma: If $\{u_n\}$ and $\{v_n\}$ are sequences of positive reals and $$\lim_{n → ∞} u_n = u > 0,\quad \varlimsup_{n → ∞} v_n = v > 0,$$
  then $\varliminf\limits_{n → ∞} \dfrac{u_n}{v_n} = \dfrac{u}{v}$.

Proof: On the one hand, since $\varlimsup\limits_{n → ∞} v_n = v$, there exists a subsequence $\{v_{n_k}\}$ such that $\lim\limits_{n → ∞} v_{n_k} = v$, then$$
\varliminf_{n → ∞} \frac{u_n}{v_n} = \lim_{m → ∞} \inf_{n \geqslant m} \frac{u_n}{v_n} \leqslant \lim_{m → ∞} \inf_{k \geqslant m} \frac{u_{n_k}}{v_{n_k}} = \varliminf_{k → ∞} \frac{u_{n_k}}{v_{n_k}} = \lim_{k → ∞} \frac{u_{n_k}}{v_{n_k}} = \frac{u}{v}.
$$
On the other hand, for any $0 < ε < 1$, there exists $N \geqslant 1$ such that $u_n > (1 - ε)u$ for $n \geqslant N$. Thus for $m \geqslant N$,$$
\inf_{n \geqslant m} \frac{u_n}{v_n} \geqslant \inf_{n \geqslant m} \frac{(1 - ε)u}{v_n} = \frac{(1 - ε)u}{\sup\limits_{n \geqslant m} v_n},
$$
which implies$$
\varliminf_{n → ∞} \frac{u_n}{v_n} = \lim_{m → ∞} \inf_{n \geqslant m} \frac{u_n}{v_n} \geqslant (1 - ε)u \lim_{m → ∞} \frac{1}{\sup\limits_{n \geqslant m} v_n} = \frac{(1 - ε)u}{\lim\limits_{m → ∞} \sup\limits_{n \geqslant m} v_n} = (1 - ε) \frac{u}{v},
$$
and making $ε → 0^+$ yields $\varliminf\limits_{n → ∞} \dfrac{u_n}{v_n} \geqslant \dfrac{u}{v}$. Therefore, $\varliminf\limits_{n → ∞} \dfrac{u_n}{v_n} = \dfrac{u}{v}$.

Now return to the question. Define $a_n = -g_n$, then $a_{n + 1} = \dfrac{1}{a_n + c_n}$ and $a_1 = c_1$. In fact, a more general proposition can be proved.

Proposition: Suppose that $\{b_n\}$ and $\{c_n\}$ are sequences of positive reals and$$
\lim_{n → ∞} b_n = b > 0,\quad \lim_{n → ∞} c_n = c > 0.
$$
  If $\{a_n\}$ satisfies $a_1 > 0$ and $a_{n + 1} = \dfrac{b_n}{a_n + c_n}$ for $n \geqslant 1$, then $\lim\limits_{n → ∞} a_n$ exists.

Proof: First, it is easy to see that $a_n > 0$ for each $n$. Since there exists $N \geqslant 1$ such that$$
\frac{b}{2} < b_n < 2b, \quad \frac{c}{2} < c_n < 2c, \quad \forall n > N
$$
then for $n > N$,$$
a_n < \frac{b_n}{c_n} < \frac{4b}{c} \Longrightarrow a_n > \frac{b_n}{\dfrac{4b}{c} + c_n} > \frac{b}{\dfrac{8b}{c} + 4c},
$$
which implies that$$
l := \varliminf_{n → ∞} a_n \geqslant \frac{b}{\dfrac{8b}{c} + 4c} > 0, \quad L := \varlimsup_{n → ∞} a_n \leqslant \frac{4b}{c} < +∞.
$$
Now, making $n → ∞$ in $a_{n + 1} = \dfrac{b_n}{a_n + c_n}$ yields$$
l = \varliminf_{n → ∞} a_{n + 1} = \frac{\lim\limits_{n → ∞} b_n}{\varlimsup\limits_{n → ∞} a_n + \lim\limits_{n → ∞} c_n} = \frac{b}{L + c},
$$
and analogously $L = \dfrac{b}{l + c}$, thus$$
l = \frac{b}{c + \dfrac{b}{c + l}} \Longrightarrow l^2 + cl - b = 0 \Longrightarrow l = \frac{1}{2} (-c + \sqrt{c^2 + 4b}),
$$
and analogously$$
L^2 + cL - b = 0 \Longrightarrow L = \frac{1}{2} (-c + \sqrt{c^2 + 4b}).
$$
Therefore, $l = L$, which implies that $\lim\limits_{n → ∞} a_n$ exists and is finite.
