The q -> E^(-2 Pi) special case of Sum[DivisorSigma[1,2*n+1] q^n,{n,0,∞}] is way simpler than any of the three eta'/eta log derivatives that comprise it. So I went after the q->E^(-4Pi/3) case: 1 + 2 n Sum[--------------------------------------, {n, 0, ∞}] -> (4 n Pi)/3 -(4/3) (1 + 2 n) Pi E (1 - E ) 1/4 2 (2 Pi)/3 1 4 Sqrt[3] (1 - Sqrt[2] + 3 + Sqrt[3]) E Gamma[-] 4 ----------------------------------------------------------- 3 512 Pi (The quadrinomial optionally factors into binomials. 1 - Sqrt[2] + 3^(1/4) + Sqrt[3] == (1 + 3^(1/4)) (Sqrt[2] + 3^(1/4))/ ((1 + Sqrt[2]) (-1 + 3^(1/4)) (Sqrt[2] + Sqrt[3])). ) But I was not able to derive this sum from the three eta'/eta quoted in the Sum[DivisorSigma[1,2*n+1] q^n,{n,0,∞}] below. I had eta'/eta (i/3) and (2i/3), but to get (4i/3) I had to differentiate Jacobi's Aequatio, which invoked eta'(i/3), eta'(2i3), eta(i/3), eta(2i/3), and eta(4i/3). This created a truly intractable surdstorm. It was not even possible to numerically circumvent RootReduce and reconstruct a simplification with RootApproximant. So I simply RootApproximated the above rhs, from which I reconstructed DedekindEta'[(4 I)/3]==(1/(\[Pi]^(11/4)))((3 I 3^(1/8) (-1+3^(1/4))^2 Sqrt[1+3^(1/4)] \[Pi]^2 Gamma[1/4])/(32 2^(2/3) (-1+3^(1/8))^2 (1+3^(1/8))^2 (Sqrt[2]-3^(1/4))^(3/8) (1+Sqrt[3])^(13/24) (2+Sqrt[2]+Sqrt[3]+Sqrt[6])^(1/4))+(I (-63-6 Sqrt[2]+27 3^(1/4)+18 Sqrt[2] 3^(1/4)-9 Sqrt[3]+21 3^(3/4)+14 Sqrt[2] 3^(3/4)-30 Sqrt[6]) Gamma[1/4]^5)/(1024 2^(2/3) 3^(7/8) (-1+3^(1/8))^2 (1+3^(1/8))^2 (Sqrt[2]-3^(1/4))^(3/8) Sqrt[1+3^(1/4)] (1+Sqrt[3])^(13/24) (2+Sqrt[2]+Sqrt[3]+Sqrt[6])^(1/4))) --rwg Speaking of the Aequatio, Mike Hirschhorn has come up with two nifty derivations and a striking observation: gosper.org/billg102615.pdf For 𝜙 and ψ see the earlier draft gosper.org/billg102115-2.pdf --rwg On Wed, Oct 21, 2015 at 3:15 AM, Bill Gosper <billgosper@gmail.com> wrote:

but I sure missed it. Pi 1 4 2 E (-)! 1 + 2 n 4 Sum[------------------------------, {n, 0, oo}] == ----------- 2 n Pi -2 (1 + 2 n) Pi 3 E (1 - E ) Pi

I got this as the q -> E^(-2 Pi) special case of Sum[DivisorSigma[1,2*n+1] q^n,{n,0,∞}]== Sum[(1 + 2 n) q^n/(1 - q^(1 + 2 n)), {n, 0, ∞}] == ( I (Derivative[1][DedekindEta][-((I Log[q])/(4 \[Pi]))]/ DedekindEta[-((I Log[q])/(4 \[Pi]))] - ( 3 Derivative[1][DedekindEta][-((I Log[q])/(2 \[Pi]))])/ DedekindEta[-((I Log[q])/(2 \[Pi]))] + ( 2 Derivative[1][DedekindEta][-((I Log[q])/\[Pi])])/ DedekindEta[-((I Log[q])/\[Pi])]))/(2 \[Pi] Sqrt[q])

As suggested by Sunday's "notational modularity" item, I have a bunch more special values of eta and eta' to crank into this. (I bet you can hardly wait.) Also, eta[q], eta[q^2], and eta[q^4] are polynomially related by Jacobi's aequatio identica satis abstrusa <https://www.google.com/search?client=safari&rls=en&q=aequatio+identica+satis+abstrusa&ie=UTF-8&oe=UTF-8> , which should have an eta' analog. --rwg Vaguely related: ArcLength[{t, Sqrt[Cos[t]]}, {t, -π/2, π/2}] == Integrate[Sqrt[1 + (1/4)*Sin[t]*Tan[t]], {t, -Pi/2, Pi/2}] but the ISC says HUH? Really? (Connection: Area under that arc = π^(3/2)/(2 Sqrt[2] (1/4)!^2). )