Estimate on Limit of Recursive Sequence

Question

How can I estimate (via a lower bound) the limit of the recursive sequence $$P_{n+1}=P_n-\frac{C(P_n-1)^2}{(2^n+C)(P_n+C2^{-n})}$$ where $0<C<1$ and $1<P_0<2$. Let $P_{\infty}=\lim_{n\to\infty}P_n$. The best estimate I can get is $$\begin{aligned} P_{\infty}-P_0 &=-\sum_{n=0}^{\infty}\frac{C(P_n-1)^2}{(2^n+C)(P_n+C2^{-n})} \\ &> -\sum_{n=0}^{\infty}\frac{C(P_n-1)^2}{2^nP_n} \\ &> -\frac{(P_0-1)^2}{P_0}\sum_{n=0}^{\infty}\frac{C}{2^n} \end{aligned}$$ where from line 2 to 3 I use the fact that $(P_n)$ is decreasing. Hence $$P_{\infty}> P_0-2C\frac{(P_0-1)^2}{P_0}$$ This is a fairly decent bound when looking at the absolute difference, but for my application I need as much precision as possible so I'm hoping to find a better lower bound. Special functions are acceptable if they arise.

Edit:
As per Wolfram, we have that $$\begin{aligned} &\sum_{n=0}^{\infty}\frac{C(P_0-1)^2}{(2^n+C)(P_0+C2^{-n})} \\ =&\frac{P_0-1}{\ln(2)}\Big(\psi_2(\log_2(-1/C))-\psi_2(\log_2(-P_0/C))\Big) \end{aligned}$$ which gives $$P_{\infty}>P_0-\frac{P_0-1}{\ln(2)}\Big(\psi_2(\log_2(-1/C))-\psi_2(\log_2(-P_0/C))\Big)$$

Answer 1

If it has a limit, it is a fixed point of your recurrence, i.e. a solution to $$P = P − \frac{C (P − 1)^2}{(2^n + C) (P + C \cdot 2^{−n})}$$ This is a quadratic equation.