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we have the sequence of positive real numbers ${a_{j}}$ , such that : $$ \frac{1}{j+1}<a_{j}<\frac{1}{j} \;\;\;\;,\;j\geqslant 1$$ and we have, for every integer $n\geqslant 1$, the equality :

$$\frac{n+2}{n+1}=\prod_{j=1}^{\infty}\left(1-\frac{a_{j}^{2}}{(a_{j}n+a_{j}-1)^{2}}\right)$$ furthermore, the infinite product : $$\prod_{j=1}^{\infty}(1+a_{j})e^{-a_{j}}$$ is convergent. in fact, there is an entire function defined as: $$f(x)=C\prod_{j=1}^{\infty}(1+xa_{j})e^{-xa_{j}}$$ such that, at negative integers: $$f(-n)=K(-1)^{n}n!$$ $C$ and $K$ being constants. can we prove that such a sequence exists ? how can we solve for the numbers $a_{j}$ ?

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If we define $b_j=1/{a_j}$ then we have $j<b_j<j+1$ for $j \geq{1}$ and the product becomes $$A=\prod_{j=1}^{\infty} \frac{(n+2-b_j)(n-b_j)}{(n+1-b_j)^2}.$$ I guess the idea is to consider the finite product for $j=1,\cdots,m$ say and cancel out as many terms as possible. That can be achieved if $$b_j=b_{j-1}+1, j=2,3,\cdots$$ but since none of the $b_j$ is an integer, no factor is of the form $n$ plus an integer. Instead we get $$A=\frac{n+2-b_1}{n+1-b_1}$$ Are you sure you wrote everything down correctly?

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