Closed form for $\sum_{n=0}^\infty\frac{\operatorname{Li}_{1/2}\left(-2^{-2^{-n}}\right)}{\sqrt{2^n}}$

Solution 1:

Result: $$\boxed{\; S=\sqrt{2}\,\operatorname{Li}_{1/2}\left(\frac14\right)-\sqrt{\displaystyle\frac{\pi}{\ln 2}}\;}\tag{1}$$

Derivation:

1. Let us denote $$a_n=\frac{\operatorname{Li}_{1/2}\left(-2^{-2^{-n}}\right)}{2^{n/2}}, \qquad b_n=\frac{\operatorname{Li}_{1/2}\left(2^{-2^{-n}}\right)}{2^{n/2}}.$$ These quantities satisfy the recurrence relation $$a_n+b_n=b_{n-1},\tag{2}$$ which follows from the identity $$\sqrt{2}\,\operatorname{Li}_{1/2}\left(z^2\right)= \operatorname{Li}_{1/2}\left(z\right)+\operatorname{Li}_{1/2}\left(-z\right).$$

2. The relation (2) implies that $$b_N+\sum_{n=0}^Na_n=b_{-1},$$ and therefore our series telescopes: $$S=\sum_{n=0}^{\infty}a_n=b_{-1}-b_{\infty}.\tag{3}$$

3. The polylogarithm asymptotics $$\operatorname{Li}_{1/2}(x)\sim\frac{\sqrt{\pi}}{\sqrt{1-x}}\quad \text{as}\;\; x\rightarrow 1^-,$$ implies that $\displaystyle b_{\infty}=\sqrt{\frac{\pi}{\ln 2}}$. Being combined with (3), this finally gives the above answer (which is further confirmed numerically).