Is there any formula for $S(k)=\sum_{n=1}^{\infty} \frac{1}{n^{k} k^{n}},$ where $k\in N$? I had just used the definite integral $$\int_{0}^{\frac{1}{2}} \frac{\ln (1-x)}{x}  d x $$  to evaluate the infinite sum
$$ \displaystyle S=\sum_{n=1}^{\infty} \frac{1}{n^{2} 2^{n}}. $$
However, I just wonder if there is a formula for a more general series $$ \displaystyle S(k)=\sum_{n=1}^{\infty} \frac{1}{n^{k} k^{n}}, \text{ where }k \in N. $$
Now let me start with the former using the infinite geometric series
$ \displaystyle \frac{1}{1-t}=\sum_{n=0}^{\infty} t^{n} \quad \text { for }|t|<1 \tag*{(1)}$
Integrating both sides of (1) w.r.t. t from 0 to x yields
$ \displaystyle \begin{array}{r} \displaystyle \int_{0}^{x} \frac{1}{1-t} d t= \displaystyle \sum_{n=0}^{\infty} \int_{0}^{x} t^{n} d t \\ \displaystyle -\ln (1-x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n}\end{array} \tag*{(2)}$
Dividing the equation (2) by x and integrating both sides from $0$ to $\frac{1}{2} $ gives
$ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2} 2^{n}}=\displaystyle -\int_{0}^{\frac{1}{2}} \frac{\ln (1-x) d x}{x} \tag*{(3)}$
Now let’s tackle the integral
$ \displaystyle J:=\int_{0}^{\frac{1}{2}} \frac{\ln (1-x) d x}{x} \tag*{} $
by the substitution $y=\ln (1-x) $.
$\displaystyle  \begin{aligned}J &=\int_{-\ln 2}^{0} \frac{y}{1-e^{y}} e^{y} d y \\&=\int_{-\ln 2}^{0} y \sum_{k=0}^{\infty} e^{(k+1) y} d y \\&=\sum_{k=0}^{\infty} \int_{-\ln 2}^{0} y e^{(k+1) y} d y\\ &\stackrel{I B P}{=} \sum_{k=0}^{\infty} \frac{1}{k+1}\left(\ln 2 \cdot \frac{1}{2^{k+1}}-\left[\frac{e^{(k+1) y}}{k+1}\right]_{-\ln 2}^{0}\right) \\&= \ln 2 \sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}-\sum_{k=1}^{\infty}\frac{1}{k^{2}}+\sum_{k=1}^{\infty} \frac{1}{k^{2} 2^{k}} \\&= \ln 2 \sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}-\frac{\pi^{2}}{6}+S\end{aligned} \tag*{} $
$ \displaystyle \textrm{Putting }x= \dfrac{1}{2} \textrm{ in (2) yields }$
$\displaystyle  \sum_{k=1}^{\infty} \frac{1}{k \cdot 2^{k}}=-\ln 2 \tag*{} $
Now we can conclude that
$ \displaystyle \begin{array}{c}\displaystyle S=-\left(\ln ^{2} 2-\frac{\pi^{2}}{6}+S \right) \\\boxed{S(2)=\sum_{n=1}^{\infty} \frac{1}{n^{2} 2^{n}} =\frac{1}{2}\left(\frac{\pi^{2}}{6}-\ln ^{2} 2\right)}\end{array}\tag*{} $
How about the general series $S(k)$?
:|D Wish you enjoy the proof!
Your suggestion, comments and formula for evaluating the general series S(k) are warmly welcome!
 A: As already pointed out by @Varun Vejalla in the comments section, the polylogarithm function is the way to go.
It´s defined by
$$\text{Li}_n\left(x\right)=\sum_{k=1}^\infty \frac{x^k}{k^n} \tag{1}$$
Letting $n=2$ and $x=\frac12$ in $(1)$ we obtain
$$\text{Li}_2\left(\frac12\right)=\sum_{k=1}^\infty \frac{1}{k^2 2^k} \tag{2}$$
Now, there is a functional equation that the dilogarithm function obbeys, namelly:
$$\text{Li}_2\left(1-\frac1x \right)=-\frac{\ln^2\left(x\right)}{2}-\text{Li}_2\left(1-x\right) \tag{3}$$
Letting $x=\frac12$ in $(3)$
$$\text{Li}_2\left(-1 \right)=-\frac{\ln^2\left(\frac12\right)}{2}-\text{Li}_2\left(\frac12\right)$$
which gives us
$$\text{Li}_2\left(\frac12 \right)=\frac{\pi^2}{12}-\frac{\ln^2\left(2\right)}{2} \tag{4}$$
euqting  $(2)$ and $(4)$ we have
$$\sum_{k=1}^\infty \frac{1}{k^2 2^k}=\frac{\pi^2}{12}-\frac{\ln^2\left(2\right)}{2} \qquad \blacksquare$$
Which is the desired result. We used the fact that
$$\text{Li}_2\left(-1\right)=\sum_{k=1}^\infty \frac{(-1)^k}{k^2}=-\eta(2)=-\frac{\pi^2}{12}$$

We can go further with this method. Recall the Trilogarithm function
$$
\text{Li}_{3}(x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{3}}
\tag{5}$$
letting $x=\frac{1}{2}$ in $(5)$ we obtain
$$
\operatorname{Li}_{3}\left(\frac{1}{2}\right)=\sum_{n=1}^{\infty} \frac{1}{2^{n} n^{3}}
\tag{6}$$
The Trilogarithm has the following identity
$$
L i_{3}(x)+L i_{3}(1-x)+L i_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2}
\tag{7}$$
plugging $x=\frac{1}{2}$ in $(7)$ we obtain
$$
\begin{gathered}
L i_{3}\left(\frac{1}{2}\right)+L i_{3}\left(1-\frac{1}{2}\right)+L i_{3}\left(1-\frac{1}{2}\right)=\zeta(3)+\frac{\ln ^{3}\left(\frac{1}{2}\right)}{6}+\frac{\pi^{2} \ln \left(\frac{1}{2}\right)}{6}-\frac{\ln ^{2}\left(\frac{1}{2}\right) \ln \left(1-\frac{1}{2}\right)}{2} \\
2 L i_{3}\left(\frac{1}{2}\right)+L i_{3}(-1)=\zeta(3)-\frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{6}+\frac{\ln ^{3}(2)}{2}
\end{gathered}
$$
Note that setting $x=-1$ in $(5)$ we get that $\text{Li}_{3}(-1)=-\eta(3)=-\frac{3}{4} \zeta(3)$, then
$$
\begin{gathered}
2 L i_{3}\left(\frac{1}{2}\right)=\frac{3}{4} \zeta(3)+\zeta(3)+\frac{\ln ^{3}(2)}{3}-\frac{\pi^{2} \ln (2)}{6} \\
L i_{3}\left(\frac{1}{2}\right)=\frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{12}+\frac{7}{8} \zeta(3)
\end{gathered}
$$
And we finally conclude that
$$
\sum_{n=1}^{\infty} \frac{1}{2^{n} n^{3}}=\frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{12}+\frac{7}{8} \zeta(3) \qquad \blacksquare
$$
Of course higher functional equations exist for higher polylog functions and the same methodology could be apllied,  but these equations get more and more complex.
