The completed identity (s > 2)
σs−1(1)e−π√3 − σs−1(2)e−2π√3 + σs−1(3)e−3π√3 − ⋯
= −2 Γ(s)ζ(s)/(2π)s · { cos(πs/6) + 2cos(π(s+1)/6)cos(π(s−1)/6) · R(s) },
R(s) = Σμ,ν≥1, (μ,ν)=1 cos( s·tan−1( (μ−ν) / ((μ+ν)√3) ) ) / (μ²+μν+ν²)s/2
Source: G. E. Andrews, B. C. Berndt, Ramanujan's Lost Notebook, Part IV (Springer 2013), §9.2–9.3, pp. 215–216 (lost notebook pp. 270–271). There K/(λ√3) appears with K, λ undefined. Proof: regroup the Andrews–Berndt form over the six units of the Eisenstein integers; three rotations of each primitive point fall in the half-plane μ ≥ 1 and combine into 4cos(π(s+1)/6)cos(π(s−1)/6)cos(sψ) with tan ψ = (μ−ν)/(√3(μ+ν)).