Does the following sum $$ S = \sum_{n\geq 2}(-1)^n \mathrm{Li}_2(2/n) = 1.14434\ 42096\ 91982\ 23727\ 39852\ 45805\ldots $$ have a closed form in terms of known constants?

Neither the inverse symbolic calculator nor wolfram alpha could suggest anything for it.

There is also the related sum $$ S_2 = \sum_{n\geq 2}\left(\mathrm{Li}_2(2/n) - 2/n\right) = 1.14135\ 80945\ 90055\ 78983\ 33729\ 08670\ldots $$ to which I would like to know a closed form.

Also, this one: $$ S_3 = \sum_{n\geq 2}(-1)^n\mathrm{Li}_2(4/n^2) = 1.30537\ 19631\ 37203\ 80215\ 02160\ 56689\ldots $$

  • $\begingroup$ I would be extremely pleased if that has a closed form. It would blow me away. $\endgroup$ Nov 6, 2014 at 2:28
  • $\begingroup$ Do you have any reason to think that a closed form exists ? $\endgroup$ Nov 6, 2014 at 5:05
  • $\begingroup$ @ClaudeLeibovici It might; if you expand the dilogarithm in power series and change the order of summation you get a sum of zeta functions; those sometimes have closed forms, but I couldn't find one here. $\endgroup$
    – Kirill
    Nov 6, 2014 at 18:27

1 Answer 1


$$\sum_{n\geq 2}(-1)^n \,\text{Li}_2\!\left(\frac{2}{n}\right)=\sum_{n\geq 2}(-1)^n \sum_{m\geq 1}\frac{2^m}{n^m m^2}=\sum_{m\geq 1}\frac{2^m(\eta(m)-1)}{m^2}\\=\sum_{m\geq 1}\frac{2^m(\zeta(m)-1)-2\zeta(m)}{m^2}\\=\sum_{m\geq 2}\frac{(2^m-2)(\zeta(m)-1)}{m^2}+\frac{\pi^2}{3}-2$$ and: $$ \sum_{m\geq 2}\frac{\zeta(m)-1}{m}\,x^m = (1-\gamma)x+\log\Gamma(-x)$$ comes from the Weierstrass product for the $\Gamma$ function. The last function has an elementary integral over any interval of the form $(a,a+1)$ by Raabe's theorem, but that does not hold if we multiply such function by $\frac{1}{x}$, so it is unlikely that your series have a nice closed form.


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