Limit of $p_n = f(a, n) + (1- f(a, n))p_{n - 1}, p_0 = a, 0 < a < 0.5$

Question

How does one compute the limit $\lim\limits_{n \rightarrow +\infty} p_n$ for the sequence $p_n = f(a, n) + (1- f(a, n))p_{n - 1}$ with $p_0 = a$, where $0 < a < \dfrac{1}{2}$, $f(a,n)=\dfrac{a}{2^n}$?

Answer 1

I worked on your problem and at last found the same thing that Kelenner said in his comment, anyway I write it to be another comment.

$p_n = p_{n - 1} + (1 - p_{n - 1})\dfrac{a}{2^n}=\dfrac{a}{2^n}+(1-\dfrac{a}{2^n})p_{n-1}=\lambda_n+(1-\lambda_n)p_{n-1}$ where $\lambda_n=\dfrac{a}{2^n}>0$.

$p_n =\lambda_n+(1-\lambda_n)p_{n-1}$ shows $a=p_0< p_1< p_2< p_3<\cdots<1$ then it's converge. Also $$\frac{1-p_n}{1-p_{n-1}}=1-\lambda_n$$ and $$\frac{1-p_k}{1-a}=\prod\limits_{n=0}^k\frac{1-p_n}{1-p_{n-1}}=\prod\limits_{n=0}^k(1-\lambda_n)$$ Thus $$\lim_{k\to\infty}p_k=1-(1-a)\prod\limits_{n=0}^\infty(1-\dfrac{a}{2^n})$$ right product has simplified by Jack D'Aurizio here.