Given $ q\in\left(0,\frac{1}{2}\right) $, and $ r_{i}=\frac{1+i^{-q}}{2},t_{i}=1-r_{i}$. Let $A_{n}$ be the set of all possible sequences of $n$ ones and $n$ zeroes (there are ${2n \choose n}$ such sequences). We define for such sequence $a\in A_{n}$ $$ z_{i,a}=\begin{cases} r_{i} & a_{i}=0\\ t_{i} & a_{i}=1 \end{cases} $$ and then we define the sequence $ b_{n} $ as follows: $$ b_{n}=\sum_{a\in A_{n}}\left(\prod_{i=1}^{2n}z_{i,a}\right) $$ The series $ \sum b_{n} $ converges, and I want to prove it.
I tried to do it by bounding $ b_{n} $ from above, and then using the comparison test, but I failed. We know that $ \left|A_{n}\right|={2n \choose n}$, and we also can consider only $ a\in A_{n} $ with $ a_{1}=0 $ (since otherwise we will have $ z_{1,a}=t_{1}=0 $, which will make the product be zero). We also know that $ r_{i} $ is decreasing and $ t_{i} $ is increasing. The naive bound I got is $$ b_{n}\leq{2n-1 \choose n-1}\left(\prod_{i=1}^{n}r_{i}\right)\left(\prod_{i=n+1}^{2n}t_{i}\right) $$ but the RHS doesn't even converge to $ 0 $, so this attempt failed.