Asymptotic expansion of $\operatorname{Li}_{-x}(1/2)$ 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.
 A: 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.
A: 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 $3.55\times 10^{-15}$.
