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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 $a_n\to 1$?

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A limit exists. Notice that $$\begin{eqnarray}\frac{1+\exp(-2n e^{-|j|/n})}{1+\exp(-(1+e^{-1/n})n e^{-|j|/n})}<1&\iff& 1+\exp(-2n e^{-|j|/n})<1+\exp(-(1+e^{-1/n})n e^{-|j|/n})\\ &\iff& -2n e^{-|j|/n}<-(1+e^{-1/n})n e^{-|j|/n}\\ &\iff& 2>1+e^{-1/n}\\ &\iff& 1>e^{-1/n}\\ \end{eqnarray}$$ which is true for all $n>0$, thus $a_n<1$ for all $n>0$. But the sequence should be eventually monotonic. –  Alex Becker Mar 9 '12 at 9:04
    
How do you prove that the sequence $a_n$ is eventually monotonic? It is true that each term is eventually monotonic in $n$, but the place when it starts to be monotonic depends on $j$ and grows when $j$ grows, so I don't see how this implies monotonicity of $a_n$. –  user26565 Mar 9 '12 at 13:58
    
My "answer" below is indeed wrong. Maybe an idea: in fact we can take the product over $j\geq 1$. For a fixed $n$, we can write $j=q_nn+r_n$ where $0\leq r_n<n-1$. Then we can bound $\exp\left(-\frac{r_n}n\right)$ to get an above bound for $a_n$. But I will do the computations carefully, since it may not work. –  Davide Giraudo Mar 11 '12 at 11:19
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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 $1$.

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I have some problems with properly using align environment, but I cannot figure out how to correct it. –  user26565 Mar 12 '12 at 1:59
    
The MathJax implementation here is somewhat idiosyncratic. To get line breaking to work in align environments, you need to either double the backslashes (\\\\ instead of \\) or wrap the whole \begin{align}...\end{align} in double dollar signs. See this meta question: Is Mathjax supposed to be 100% compatible with Latex? –  Rahul Mar 12 '12 at 2:41
    
Thank you Rahul! –  user26565 Mar 12 '12 at 2:46
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