Non vanishing of an infinite product

Question

I need to prove that the infinite product $$\prod_n \left(1-\frac{1} {(a^n+1)^2} \right)^{\frac{a^n}{n}} $$ with $a$ an integer $\geq 3$, converges to a real number $L$ such that $0<L<1$.

It's immediate to see that $L<1$, as we have $\left(1-\frac{1} {(a^n+1)^2} \right)^{\frac{a^n}{n}}<1$, but I didn't found a way to say that the limit is not zero.

Thanks for your help.

Answer 1

Since for any $z\in(0,1)$ we have: $$\log(1-z)\geq\frac{z}{z-1}$$ it is sufficient to prove that $$ \sum_{n\geq 1}\frac{a^n}{n}\frac{\frac{1}{(a^n+1)^2}}{\frac{1}{(a^n+1)^2}-1}=-\sum_{n\geq 1}\frac{1}{n}\cdot\frac{1}{a^n+2} $$ is converging, but since $a\geq 3$ we have: $$0\leq \sum_{n\geq 1}\frac{1}{n}\cdot\frac{1}{a^n+2}\leq\sum_{n\geq 1}\frac{1}{n 3^n}\leq\log\frac{3}{2}$$ hence: $$\prod_{n\geq 1}\left(1-\frac{1}{(a^n+1)^2}\right)^{\frac{a^n}{n}}\geq\frac{2}{3}.$$

Answer 2

Since $a\ge3$, $$ \frac{1} {(a^n+1)^2}\le\frac{1}{16}\quad\forall n\ge1. $$ Let $C>0$ be such that $\log(1-x)\ge-C\,x$ if $0<x\le1/16$. For any $N>1$

$$\begin{align} \log\Biggl(\prod_{n=1}^N \Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)^{\frac{a^n}{n}}\Biggr)&=\sum_{n=1}^N\frac{a^n}{n}\log\Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)\\ &\ge-C\sum_{n=1}^N\frac{a^n}{n}\,\frac{1}{(a^n+1)^2}\\ &\ge-C\sum_{n=1}^N\frac{1}{a^n}\\ &=-\frac{C}{a-1}. \end{align}$$ Thus $$ \prod_{n=1}^N \Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)^{\frac{a^n}{n}}\ge e^{-\tfrac{C}{a-1}}>0. $$