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 serie $\sum_{k=0}^{+ \infty}(1-|L_{n}^{5}(3)|)^{k}$ is convergent, i.e
$$\sum_{k=0}^{+ \infty}(1-|L_{n}^{5}(3)|)^{k} < + \infty$$
I would like to knwo 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 constante 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.