Well A last conjecture :
Let :
$$f\left(x\right)=\left(\left|\int_{0}^{x}\prod_{n=1}^{\infty}\left(1+e^{-t-n}\right)dt-x-1\right|\right)^{\frac{1}{x}}$$
Then it seems we have :
$$\prod_{n=1}^{\infty}\left(1+e^{-n}\right)+\lim_{x\to 0}\ln\left(f\left(x\right)\right)=1$$
Now I think it's not (so) trivial and I cannot solve it.I really guess there is a general result behind this but cannot find it .I have also tried L'hopital rule without a success .
How to (dis)prove it ?
By Taylor's theorem $$ \int_0^x {\prod\limits_{n = 1}^\infty {(1 + {\rm e}^{ - t - n} )} {\rm d}t} = \bigg( {\prod\limits_{n = 1}^\infty {(1 + {\rm e}^{ - n} )} } \bigg)x + \mathcal{O}(x^2 ) $$ as $x\to 0$. Then \begin{align*} \log f(x) = \frac{1}{x}\log\! \bigg( {1 + \bigg( {1 - \prod\limits_{n = 1}^\infty {(1 + {\rm e}^{ - n} )} } \bigg)x + \mathcal{O}(x^2 )} \bigg) = \bigg( {1 - \prod\limits_{n = 1}^\infty {(1 + {\rm e}^{ - n} )} } \bigg) + \mathcal{O}(x) \end{align*} as $x\to 0$. Your claim follows.
