Infinite sum: $\sum_{x=1}^\infty\frac{\ln x}{e^x}$

Question

I have tried to find this infinite sum all over the place but could not find it. Does anyone know a method for finding this sum: $$\lim_{n \to \infty} \sum_{x=1}^n {\ln(x)\over e^x}$$

Any help is greatly appreciated

Answer 1

$$S=\sum_{k=1}^\infty \frac{\ln(k)}{e^k}$$ It is easy to prove the convergence. That is not the question.

The question is about the existence, or not, of a closed form. In other words, if $S$ is related to a standard special function.

Consider the kind of special functions called "Polylogarithm" : $$\text{Li}_\nu(z)=\sum_{k=1}^\infty \frac{z^k}{k^\nu}=\sum_{k=1}^\infty e^{-\nu\,\ln(k)}z^k$$ $$\frac{\partial }{\partial \nu}\text{Li}_\nu(z)=-\sum_{k=1}^\infty \ln(k)e^{-\nu\,\ln(k)}z^k=-\sum_{k=1}^\infty \frac{\ln(k)}{k^\nu}z^k$$ In the particular case $\quad\nu=0\quad\text{and}\quad z=\frac{1}{e}\quad$ this leads to $\quad \frac{\partial }{\partial \nu}\text{Li}_\nu(z)=-\sum_{k=1}^\infty \frac{\ln(k)}{e^{k}}$ $$\sum_{k=1}^\infty \frac{\ln(k)}{e^k}=-\left(\frac{\partial }{\partial \nu}\text{Li}_\nu(z)\right)\left(\nu=0\:,\:z=1/e\right)$$ This special function is implemented in WolframAlpha as $\text{PolyLog}^{(1\,,\,0)}(\nu,z)$. The first exponent means the partial derivative with regard to $\nu$. The second exponent means the partial derivative with regard to $z$, which in the present case is of degree $0$, that is the function itself.

That is why WolframAlpha gives the result on the form : $$\sum_{k=1}^\infty \frac{\ln(k)}{e^k}=-\text{PolyLog}^{(1\,,\,0)}\left(0,\frac{1}{e}\right)\simeq 0.1920928...$$