Problem
Given $p_1, \sigma > 0$, consider the following recursion \begin{equation*} p_{i}=(1-L_i)p_{i-1} \qquad i=2,\dots,k \end{equation*} where \begin{equation*} L_i \triangleq \frac{p_{i-1}}{p_{i-1}+\sigma} \end{equation*} I would like to compute the final value $p_k$ as $k\to \infty$.
My attempt
To be honest, due to its non-linearity, I don't know how to analyze the current recursion. The only thing that comes to my mind is the following: regardless the values of $p_{i-1}$ and $\sigma$, we have $L_i \in[0,1]$ if $p_{i-1}$ stay non-negative. Under this conjecture, we have the two limit cases in which $L_i\triangleq 0$ and $L_i \triangleq 1$. The evolution of $p_i$ is then confined between this two cases.
- $L_i \triangleq 1$ leads to $p_i = 0$, so in this case the limit value is zero;
- $L_i \triangleq 0$ leads to $p_i = p_{i-1}$, so in this case the limit value is $p_1$.
Based on this, I would say that $p_i$ is confined between $0$ and $p_1$ for all $i$. However, my argument is far from rigorous and does not provide a clear answer to my problem.
EDIT
Since \begin{equation*} 1-L_i = \frac{\sigma}{p_{i-1}+\sigma} \end{equation*} it follows from $p_1, \sigma >0$ that $L_i\geq 0$, so that $p_i\geq 0$ for all $i$.
Hint: $$\frac1{p_i}=\frac1{p_{i-1}}+\frac1\sigma.$$