$a_{n+1}=\dfrac{a_n+c}{1-a_nc}$. Is it possible to make $a_0, a_1, \cdots, a_{i-1}$ positive and $a_{i}$ negative for $\forall i$? ($a_0>0, c>0.$)
I tried to decide the sequence $a_i$ monotone sequence. If it is an increasing one, we can't make $a_0$ to $a_{i-1}$ positive and $a_i$ negative. So, I tried to decide on the sequence as the decreasing sequence.
To make this decrease, we have to make $a_n+c<a_n(1-a_nc)$, $a_n^2c+c<0$, $a_n^2+1<0$.
This is a contradiction. So, the sequence can't be decreasing one...
Is there something else to do from here?
If $a_{n} < 0$ for $n \geqslant i \in \mathbb{N}$, then $$ a_{n+1} - a_{n} = c\,(1+a_{n}a_{n+1}) > 0, \tag{1} \label{label} $$ indicating that $a_{n}$ is monotonically increasing, starting from $i$. By Weierstrass theorem on the limit of a monotone sequence we conclude that $a_{n}$ converges to some limit $a \in \mathbb{R}$. If we take limit from both sides in \eqref{label}, then we get an statement of the form $1 + a^{2} = 0$, leading to a contradiction.