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Let $(c_n)$ be a sequence of positive numbers converging to $1$ as $ntoinfty$. That is, $c_n > 0$ for every $n$ and $limlimits_ntoinftyc_n=1$. Let $g_1=-c_1$ and define, for $n geq 1$,$$g_n+1=frac-1c_n-g_n.$$

Question: Does $limlimits_n to infty g_n$ 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 $g_infty = dfrac-11-g_infty$.

Notice also that each $g_n+1$ can be represented as the following finite continued fraction
$$g_n+1=-cfrac1c_n+cfrac1c_n-1+cfrac1ddots+cfrac1c_1-g_1,$$
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 $limlimits_n to infty g_n$ exist?

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 $varliminflimits_n → ∞ dfracu_nv_n = dfracuv$.

Proof: On the one hand, since $varlimsuplimits_n → ∞ v_n = v$, there exists a subsequence $v_n_k$ such that $limlimits_n → ∞ v_n_k = v$, then$$
varliminf_n → ∞ fracu_nv_n = lim_m → ∞ inf_n geqslant m fracu_nv_n leqslant lim_m → ∞ inf_k geqslant m fracu_n_kv_n_k = varliminf_k → ∞ fracu_n_kv_n_k = lim_k → ∞ fracu_n_kv_n_k = fracuv.
$$
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 $m geqslant N$,$$
inf_n geqslant m fracu_nv_n geqslant inf_n geqslant m frac(1 - ε)uv_n = frac(1 - ε)usuplimits_n geqslant m v_n,
$$
which implies$$
varliminf_n → ∞ fracu_nv_n = lim_m → ∞ inf_n geqslant m fracu_nv_n geqslant (1 - ε)u lim_m → ∞ frac1suplimits_n geqslant m v_n = frac(1 - ε)ulimlimits_m → ∞ suplimits_n geqslant m v_n = (1 - ε) fracuv,
$$
and making $ε → 0^+$ yields $varliminflimits_n → ∞ dfracu_nv_n geqslant dfracuv$. Therefore, $varliminflimits_n → ∞ dfracu_nv_n = dfracuv$.

Now return to the question. Define $a_n = -g_n$, then $a_n + 1 = dfrac1a_n + c_n$ 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 $a_n + 1 = dfracb_na_n + c_n$ for $n geqslant 1$, then $limlimits_n → ∞ a_n$ exists.

Proof: First, it is easy to see that $a_n > 0$ for each $n$. Since there exists $N geqslant 1$ such that$$
fracb2 < b_n < 2b, quad fracc2 < c_n < 2c, quad forall n > N
$$
then for $n > N$,$$
a_n < fracb_nc_n < frac4bc Longrightarrow a_n > fracb_ndfrac4bc + c_n > fracbdfrac8bc + 4c,
$$
which implies that$$
l := varliminf_n → ∞ a_n geqslant fracbdfrac8bc + 4c > 0, quad L := varlimsup_n → ∞ a_n leqslant frac4bc < +∞.
$$

Now, making $n → ∞$ in $a_n + 1 = dfracb_na_n + c_n$ yields$$
l = varliminf_n → ∞ a_n + 1 = fraclimlimits_n → ∞ b_nvarlimsuplimits_n → ∞ a_n + limlimits_n → ∞ c_n = fracbL + c,
$$
and analogously $L = dfracbl + c$, thus$$
l = fracbc + dfracbc + l Longrightarrow l^2 + cl - b = 0 Longrightarrow l = frac12 (-c + sqrtc^2 + 4b),
$$
and analogously$$
L^2 + cL - b = 0 Longrightarrow L = frac12 (-c + sqrtc^2 + 4b).
$$
Therefore, $l = L$, which implies that $limlimits_n → ∞ a_n$ exists and is finite.

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, $-frac12 geq g_2k geq g_infty geq g_2m+1 geq -1, |g_2n-g_2n+1| to 0$ by standard manipulations, so $g_n to g_infty$ in this case.

Then wlog assume $|c_n-1| leq frac1100$ say for all $n geq 1$ or start from a place where that happens and renumber if you wish, so in particular $c_n geq frac99100$, call $h_n$ the sequence from step above (where $c_n=1$) and note that $g_n, h_n <0, h_n leq -frac12, h_n to g_infty$ from the above. Note also that $|g_n+1| leq frac1c_n < 2$ from our assumptions.

Then using $c_n-g_n geq frac99100, 1-h_n geq frac32$ and $|g_n+1-h_n+1| = |frac(c_n-1)+(h_n-g_n)((c_n-g_n)(1-h_n)| leq fracc_n-1frac32frac99100$, it follows that $|g_n+1-h_n+1| leq frac34(|c_n-1|+|h_n-g_n|)$. 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 $d_n+1 leq frac34(q_n+ d_n)$; 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 $d_n+k+1 leq (frac34)^kd_n+(sup_pgeq nq_p)Sigma(frac3 4)^m leq 3(frac34)^k + 4(sup_pgeq nq_p)$, for any $epsilon >0$, we pick first $n_0$, s.t. $(sup_pgeq n_0q_p) < fracepsilon8$ and then $k_0$ st. $3(frac34)^k_0 < fracepsilon2$ and then $d_m < epsilon, m>n_0+k_0$, so $d_n to 0$ indeed and we are done.