Asymptotic expansion of $\operatorname{Li}_{-x}(1/2)$

Question

I want to find the asymptotic expansion of $$\operatorname{Li}_{-x}(1/2):=\sum_{n=1}^\infty\frac{n^x}{2^n}$$as $\Re x\to+\infty$.
My Attempt
Firstly, I considered series $$f(y)=\sum_{m=1}^\infty \operatorname{Li}_{-m}(1/2)y^m$$ $$=\sum_{n=1}^\infty\frac{ny}{2^n(1-ny)}.$$ It has singular points in $|y|<\varepsilon$ for all positive $\varepsilon$. Hence $\limsup_{n\to\infty}\sqrt[n]{\operatorname{Li}_{-n}(1/2)}=+\infty$. But this is all I can got.

Answer 1

The Lindelof-Wirtinger expansion for the Lerch transcendent gives $$\operatorname{Li}_{-x} \left( \frac 1 2 \right) = \Phi \!\left( \frac 1 2, -x, 0 \right) = \Gamma(x + 1) \sum_{a_k = \ln 2 + 2 \pi i k} a_k^{-x - 1}.$$ The leading term is determined by the $a_k$ for which $a_k^{-x}$ has the fastest rate of growth. This will depend on how $x$ approaches infinity. If $\arg x$ doesn't approach critical values $\phi_k$, then there is a single dominant term, which will give a uniform approximation inside a sector: $$\operatorname{Li}_{-x} \left( \frac 1 2 \right) \sim \Gamma(x + 1) (\ln 2 + 2 \pi i k)^{-x - 1}, \\ \operatorname{Re} x \to \infty \land \phi_{k - 1} + \epsilon < \arg x < \phi_k - \epsilon, \\ \epsilon > 0, \quad \phi_k = \frac \pi 2 - \arg \ln \frac {a_{k + 1}} {a_k}.$$ $k = 0$ corresponds to the sector containing the positive real axis.

Answer 2

This is sequence $A000629$ in $OEIS$.

In the comments, you could notice that they give as asymptotics $$\text{Li}_{-x}\left(\frac{1}{2}\right)\sim\frac{x!}{[\log(2)]^{(x+1)}}$$ which seems to be extremely good. For $x=100$, the relative error is $3.55\times 10^{-15}$.