* Bill Gosper <billgosper@gmail.com> [Feb 27. 2012 14:27]:
On Sun, Feb 26, 2012 at 12:32 AM, Bill Gosper <billgosper@gmail.com> wrote:
(-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}])/ (1 + a)
as a -> -1+ . (In terms of q-hypergeometric(s) in x and q).
Here's a much better answer as two Lambert series:
Limit[(-1/2 + Sum[(a^n*QPochhammer[x/a, q, n])/QPochhammer[a*x, q, n], {n, 0, Infinity}])/(1 + a), a -> -1, Direction -> -1] == 1/4 + 2*Sum[(-x)^n/(1 - q^(2*n)), {n, 1, Infinity}] + x*Sum[q^n/(1 + q^n*x), {n, 0, Infinity}]
This came from combining Heine's first and second transformations http://dlmf.nist.gov/17.6 (17.6.6, 17.6.7) to get QHypergeometricPFQ[{a, b}, {c}, q, z]== QHypergeometricPFQ[{c/b, z}, {a z}, q, b] * QPochhammer[b, q] QPochhammer[a z, q])/ (QPochhammer[c, q] QPochhammer[z, q])
This transforms the sum in the limitand into one with numerator parameter a^2, which means all its terms after the 0th will contain a removable factor of 1-a^2 = (1-a)(1+a), and the 0th minus 1/2 and then divided by a+1 contributes one of the Lambert sums as a -> -1.
[...]
Not sure this is pertinent here, but see relation (3.1) on p.604 of %\bibitem{Chan}{Song Heng Chan: % {Generalized Lambert Series Identities}, % Proceedings of the London Mathematical Society, vol.91, no.3, pp.598-622, (2005). %}%% rel.(3.1): expression for bilateral Lambert series Can send the pdf in case the paper isn't open access. IIRC there is something similar in Fine, but can only point to his relation (18.4) on p.20 right now.