Non vanishing of an infinite product 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.
 A: 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:
$$\prod_{n\geq 1}\left(1-\frac{1}{(a^n+1)^2}\right)^{\frac{a^n}{n}}\geq\frac{2}{3}.$$
A: 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
$$
\prod_{n=1}^N \Bigl(1-\frac{1}{(a^n+1)^2} \Bigr)^{\frac{a^n}{n}}\ge e^{-\tfrac{C}{a-1}}>0.
$$
