I have the following recursion of sequence $\{A_i\}_{i=1}^\infty$ and $\{B_i\}_{i=0}^\infty$
Index
1 $~~~~~~~~~~~~~~A_1=q\sigma$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~B_1=B_0-A_1B_0/\sigma$
2 $~~~~~~~~~~~~~~A_2=A_1(1-\sigma)$ $~~~~~~~~~~~~~~~~B_2=B_1-A_2B_1/\sigma$
$n$ $~~~~~~~~~~~~~A_n=A_{n-1}(1-\sigma)$ $~~~~~~~~~~~~B_n=B_{n-1}-A_nB_{n-1}/\sigma$
where $B_0=\sigma\in(0,1)$ and $q\in(0,1)$.
So, it is clearly that $A_n=q\sigma(1-\sigma)^{n-1}$. But $B_n$ is difficult to get a simple form.
Then I want to compute the following series \begin{align*} \sum_{i=1}^{\infty}(1-\sigma)^{i-1}B_{i-1}&=\frac{1}{q}\sum_{i=1}^{\infty}A_iB_{i-1}/\sigma\\ &=\frac{1}{q}(B_0-B_1+B_1-B_2+B_2-B_3+....)\\ &=\frac{1}{q}B_0=\sigma/q \end{align*}
The last step is questionable, as it is an alternating series. One can show that $B_n$ is decreasing and is bounded by $\sigma$. So if you replace $B_n$ by $\sigma$, then this whole series is 1. So the originaly series is bounded by 1. But $B_n$ does not go to zero as $n$ increases. So it fails one of the alternating series convergence test. I know that it is a sufficient condition.
Then I am stuck at this point. How to solve this series? Is this even convergent?