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

Let
$$S=\sum_{n=0}^\infty\frac{\operatorname{Li}_{1/2}\left(-2^{-2^{-n}}\right)}{\sqrt{2^n}},\tag1$$
where $\operatorname{Li}_a(z)$ is the polylogarithm. For $a=1/2$ it can be represented as
\begin{align}\operatorname{Li}_{1/2}(z)&=\sum_{k=1}^\infty\frac{z^k}{\sqrt k}\tag2\\&=\int_0^\infty\frac z{\sqrt{\pi\,x}\ \left(e^x-z\right)}\,dx.\tag3\end{align}

How to find a closed-form expression for $S$?

#### Solutions Collecting From Web of "Closed form for $\sum_{n=0}^\infty\frac{\operatorname{Li}_{1/2}\left(-2^{-2^{-n}}\right)}{\sqrt{2^n}}$"

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).