An irrationality result for a recursive construction (Q2236581)

Let \(\{q_j\}_{j-1}^\infty\) be a sequence of non-zero complex numbers. Set recursively \[ a_{k,1}=\sum_{j=k}^\infty\frac 1{q_jq_{j+1}},\tag{1} \] \(a_{k,n}=\sum_{j=k}^\infty\frac {a_{j+2n-1}}{q_jq_{j+1}}\) for (\(n\geq 2\)) and \(f_k(x)=1+\sum_{n=1}^\infty a_{k,n}x^n\). Assume that there exists a fixed integer \(m_0\) such that \(m_0q_j\) is a Gaussian integer for all \(j\in\mathbb Z^+\). Then the author proves that if there exists \(c_0>0\) such that: 1. If \(\frac 1{\mid q_j\mid}\leq \frac {c_0}j\) for all \(j\in\mathbb Z^+\) then all \(f_k(x)\) are entire functions. In addition if \(x\not= 0\) is a Gaussian rational then \(f_k(x)\not= 0\) and \(\frac {f_{k+1}(x)}{f_k(x)}\) cannot be a Gaussian rational for all \(k\in\mathbb Z^+\). 2. If the series (1) converges and \(\frac 1{\mid q_j\mid}\leq \frac {c_0}{\sqrt j}\) and \(\mid a_{k,1}\mid \leq \frac {c_0}{\sqrt k}\) for all \(j, k\in\mathbb Z^+\) then all \(f_k(x)\) have radius convergence \(\geq \frac {\sqrt 3}{c_0^24}\). In addition if \(x\not= 0\) is a Gaussian rational with \( \frac {c_0^24x}{\sqrt 3}<1\) then \(f_k(x)\not= 0\) and \(\frac {f_{k+1}(x)}{f_k(x)}\) cannot be a Gaussian rational for all \(k\in\mathbb Z^+\).