# The Cauchy product $\sum_{n=1}^\infty \frac{\log n}{e^n}= \left( 1-\frac{1}{e} \right)\sum_{n=1}^\infty\frac{\log n!}{e^n}$

I know that the Cauchy product is defined $$\left(\sum_{n=1}^\infty\frac{\log n}{e^n}\right)\left( \sum_{n=1}^\infty\frac{1}{e^n} \right)= \sum_{n=1}^\infty\sum_{k=1}^n\frac{\log k}{e^{k+n-k+1}},$$ and from this compute $$\sum_{n=1}^\infty \frac{\log n}{e^n}= \left( 1-\frac{1}{e} \right)\sum_{n=1}^\infty\frac{\log n!}{e^n}.$$

My purpose was to state a nice identity, but I have problems. My computations were by partial summation $$\sum_{n\leq x}\frac{\log n}{e^n}=[x]\frac{\log x}{e^x}-\int_1^x [t]\frac{\frac{1}{t}-\log t}{e^t}dt,$$ where the first summand in RHS vanishes as $x\to\infty$, and the integral converges. If there are no mistakes, I know write this last as the series of $$k\int_k^{k+1}\frac{\frac{1}{t}-\log t}{e^t}dt.$$

But I don't know if it is possible continue, because Wolfram Alpha say that $\sum_{n\leq x}\frac{\log n}{e^n}$ is equal to $$-PolyLog^{(1,0)}(0,\frac{1}{e}).$$

I don't know if such exercise is in the literature, I ask it

Question. Can you compute $$\sum_{n=1}^\infty\frac{\log n!}{e^n}$$ or explain what is $$-PolyLog^{(1,0)}(0,\frac{1}{e})$$ (I say a understandable explanation, how is defined this last special function, how to evaluate this particular value and why converges it) to get an identity? Thanks in advance.

The polylogarithm function is defined as $$\textrm{Li}_{s}\left(z\right)=\sum_{k\geq1}\frac{z^{k}}{k^{s}}$$ for all complex $s$ and for $\left|z\right|<1$. So observe that $$\frac{\partial}{\partial s}\left(\textrm{Li}_{s}\left(\frac{1}{e}\right)\right)=-\sum_{k\geq1}\frac{\log\left(k\right)}{e^{k}k^{s}}$$ and so $$\frac{\partial}{\partial s}\left(\textrm{Li}_{s}\left(\frac{1}{e}\right)\right)_{s=0}=-\sum_{k\geq1}\frac{\log\left(k\right)}{e^{k}}$$ and this is the $PolyLog^{\left(1,0\right)}\left(0,\frac{1}{e}\right)$ of Wolfram alpha. So the parenthesis $(1,0)$ indicates the differentiation with respect to $s$ one time and no differentiation with respect to $z$.