Let put $\alpha=5$ and $x=3$. Consider the following set given by $$M=\lbrace \; n \in N, \; \; 0 < |L_{n}^{5}(3)| < 1 \; \rbrace$$ Where $L_{n}^{\alpha}(x)$ is the generalized Laguerre polynomial defined by $$L_{n}^{\alpha}(x)=\sum_{k=0}^{n} \frac{\Gamma(\alpha+n+1)}{\Gamma(n-k+1)\Gamma(\alpha+k+1)} \frac{(-x)^{k}}{k!}=\frac{x^{-\alpha}e^{x}}{n!} \frac{d^{n}}{dx^{n}}(x^{n+\alpha}e^{-x})$$ For the sequence $(|L_{n}^{5}(3)|)_{n \in M}$. I would like to prove that the series $\sum_{k=0}^{+ \infty}(1-|L_{n}^{5}(3)|)^{k}$ is convergent, i.e $$\sum_{k=0}^{+ \infty}(1-|L_{n}^{5}(3)|)^{k} < + \infty$$ I would like to know if I could write the expression above as $$\mbox{ for some } m > 0, \sum_{k=0}^{+ \infty}(1-|L_{n}^{5}(3)|)^{k} < m$$ where $m$ is a strictly positive constant which does not depend on $n$. Please, I need some clarification on that case, because in fact geometric series are obviously convergent but the problem here is the index $n$ must not be involved in order to prove the statement in more explicit way.