Convergence of a sequence written as infinite products Let 
$$a_n=\prod_{j\in\mathbb{Z}}\frac{1+\exp(-2n e^{-|j|/n})}{1+\exp(-(1+e^{-1/n})n e^{-|j|/n})}$$
Then each term in the product goes to $1$ as $n\to\infty$. Does $a_n\to 1$?
 A: For the sake of reference I post a solution (it does not converge to $1$), based on ideas from previous answer by Davide Giraudo (especially noticing a telescoping sum was of great help).
Let 
\begin{equation*}
w_{j,n}=\frac{1+\exp(-(1+e^{-1/n})ne^{-j/n})}{1+\exp(-2ne^{-j/n})}>1
\end{equation*}
First one notices that it suffices to prove that $\prod\limits_{j\geq 1}w_{j,n}$ converges or not to $1$.
Then 
\begin{equation*}
1+\sum_{j=1}^{J}(w_{j,n}-1)\leq\prod_{j=1}^{J}w_{j,n}\leq \exp \left(\sum_{j=1}^{J}(w_{j,n}-1)\right)
\end{equation*}
Hence $\prod\limits_{j=1}^{\infty}w_{j,n}$ converges if and only if $\sum\limits_{j=1}^{\infty}(w_{j,n}-1)$ converges, but 
$$
\begin{align*}
w_{j,n}-1&=\frac{\exp(-(1+e^{-1/n})ne^{-j/n})-\exp(-2ne^{-j/n})}{1+\exp(-2ne^{-j/n})}\\
&=\frac{\exp(-ne^{-j/n})}{1+\exp(-2ne^{-j/n})}\left[\exp(-ne^{-(j+1)/n})-\exp(-ne^{-j/n})\right]\\
&\leq\left[\exp(-ne^{-(j+1)/n})-\exp(-ne^{-j/n})\right]
\end{align*}
$$
so $$\sum_{j=1}^{+\infty}(w_{j,n}-1)\leq \lim_{j\to+\infty}\left[\exp(-ne^{-j/n})-\exp(-ne^{-1/n})\right]={1-\exp(-ne^{-1/n})}$$
converges. Therefore we may write
$$
1+\sum_{j=1}^{+\infty}(w_{j,n}-1)\leq\prod_{j=1}^{+\infty}w_{j,n}\leq \exp \left(\sum_{j=1}^{+\infty}(w_{j,n}-1)\right)
$$
And thus $a_n=\prod\limits_{j=1}^{+\infty}w_{j,n}\to 1$ if and only if $b_n=\sum\limits_{j=1}^{+\infty}(w_{j,n}-1)\to 0$.
Let now 
$$
\begin{align*}
\eta_{j,n}&=\exp(-ne^{-j/n})\left[\exp(-ne^{-(j+1)/n})-\exp(-ne^{-j/n})\right],
\\
\theta_{j,n}&=\exp\left(-ne^{-(j+1)/n}\right)\left[\exp(-ne^{-(j+1)/n})-\exp(-ne^{-j/n})\right],
\\
c_n&=\sum_{j=1}^{+\infty}\eta_{j,n},
\\
d_n&=\sum_{j=1}^{+\infty}\theta_{j,n}.
\end{align*}
$$
Then we have $\frac12\eta_{j,n}\leq w_{j,n}\leq \eta_{j,n}$, thus 
$\frac{1}{2}c_n\leq b_n\leq c_n$, and
$$
\begin{align*}
c_n+d_n&=\\&\sum_{j\geq 1} \left[\exp(-ne^{-(j+1)/n})+\exp(-ne^{-j/n})\right]\left[\exp(-ne^{-(j+1)/n})-\exp(-ne^{-j/n})\right]\\
&=\sum_{j\geq 1}\left[\exp(-2ne^{-(j+1)/n})-\exp(-2ne^{-j/n})\right]\\
&=\lim_{j\to+\infty}\left[\exp(-2ne^{-j/n})-\exp(-2ne^{-1/n})\right]\\
&=\left[1-\exp(-2ne^{-1/n})\right].
\end{align*}
$$
We also compute
\begin{align*}
\frac{\theta_{j,n}}{\eta_{j,n}}=\frac{\exp(-ne^{-(j+1)/n})}{\exp(-ne^{-j/n})}=\exp\left[e^{-j/n}n\left[1-e^{-1/n}\right]\right)
\end{align*}
And since  by Lagrange theorem for some $-1/n<c<0$ we have 
\begin{align*}
0\leq n\left[1-e^{-1/n}\right]= n\left[e^0-e^{-1/n}\right]\leq n e^c(0-(-1/n))=e^c<1
\end{align*}
we obtain that for all $j\geq 1$ and $n\geq 1$
$$
1\leq \frac{\theta_{j,n}}{\eta_{j,n}}\leq e
$$
and thus $d_n\leq e c_n$.
Finally from inequalities above we have
$$
1-\exp(-2ne^{-1/n})=c_n+d_n\leq (1+e)c_n\leq 2(1+e)b_n,
$$
hence 
$$
b_n\geq \frac{1}{2(1+e)}\left[1-\exp(-2ne^{-1/n})\right]
$$
and so $$\prod_{j=1}^{\infty}w_{j,n}\geq 1+\frac{1}{2(1+e)}\left[1-\exp(-2ne^{-1/n})\right],$$ hence does not converge to $1$. 
