* Bill Gosper <billgosper@gmail.com> [Feb 28. 2012 08:14]:
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Dawk! And better still as one Lambert series!
(first:)
Sum[((-1)^k*q^((1/2)*(-1 + k)*k)*x^k*QPochhammer[q, q, -1 + k])/ (QPochhammer[-1, q, 1 + k]*QPochhammer[-x, q, k]), {k, 1, Infinity}] == (second:) Sum[(-x)^n/(1 - q^(2*n)), {n, 1, Infinity}] + (1/2)*x*Sum[q^n/(1 + q^n*x), {n, 0, Infinity}] == (third:) (-(1/2))*x*Sum[(-q)^n/(1 + q^n*x), {n, 0, Infinity}]
The equality (first) == (third) appears to be a specialization of my relation (7): \begin{equation}\label{rel:lambert-qxt-alt} % sum(n=0,N, t^n/(1-x*q^n) ); \sum_{n\geq{}0}{ \frac{t^n}{1-x\,q^{n}} } = % % sum(n=0,N, (qbin(q,q,n)) / (qbin(x,q,n+1)*qbin(t,q,n+1)) * (-x*t)^n * q^((n^2-n)/2) ); \sum_{n\geq{}0}{ \frac{ (q;q)_n }{ (x;q)_{n+1} \, (t;q)_{n+1} } \, (x\,t)^n\, q^{(n^2-n)/2} } \end{equation} The equality (second) == (third) resembles the "well known" sum(n>=1, x*q^n/(1-x*q^n) ) == -x/(1-x) + sum(n>=1, x^n/(1-q^n) ) (but I cannot morph this one into your's right now, replacing x by -x may be a good start).
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