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 $S$?

  • $\begingroup$ Why do you need a closed form for $S$? Do you expect it to have closed form? What is the motivation for your question? $\endgroup$ – user17762 Dec 2 '13 at 19:35
  • $\begingroup$ It is a contest problem. There is a closed form, but I do not know it. $\endgroup$ – Laila Podlesny Dec 2 '13 at 19:36

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


  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).


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