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.
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$\begingroup$ $\sum_{k=0}^\infty(1-|L_n^5(3)|)^k=1/|L_n^5(3)|$. So, it appears that you want to show that $L_n^5(3)$ is bounded away from $0$, right? $\endgroup$– Iosif PinelisCommented Mar 30, 2022 at 14:21
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$\begingroup$ Yes i want to prove that $1/|L_{n}^{5}(3)|$ is bounded for $n \in M$ $\endgroup$– Assinisa HamidataCommented Mar 30, 2022 at 18:21
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$\begingroup$ OK. Then you can quite a bit simplify your question: remove the series and remove the set $M$, and just say that you want to prove that $L^5_n(3)$ is bounded away from $0$. ($1/L^5_n(3)$ is bounded by $1$ for $n\notin M$.) $\endgroup$– Iosif PinelisCommented Mar 30, 2022 at 18:29
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$\begingroup$ But i want to prove that $1/|L_{n}^{5}(3)|$ is bounded for $n \in M$. Why you remove $M$? $\endgroup$– Assinisa HamidataCommented Mar 30, 2022 at 21:55
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$\begingroup$ You can leave your $M$, if you do not want to simplify your question. The explanation of why $M$ is not needed was given in the last sentence (in the parentheses) of my previous comment. $\endgroup$– Iosif PinelisCommented Mar 30, 2022 at 22:02
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