[math-fun] I wonder if this would wiggle the needle
on the milliramanujanometer: Sum[n/(-1 + E^(2*Sqrt[11]*n*Pi)), {n, 1, Infinity}] == 1/24 - 1/(8*Sqrt[11]*Pi) - ((8 - (4*6^(1/3)*(-6127 + 63*Sqrt[33]))/ (8199 + 56809*Sqrt[33])^(2/3) + (99 + 205*Sqrt[33])/(6*(8199 + 56809*Sqrt[33]))^ (1/3))*Gamma[1/22]*Gamma[5/22]*Gamma[3/11]* Gamma[15/22])/(2112*2^(8/11)*Sqrt[11]*Pi^2* Gamma[7/22]*Gamma[10/11]) --rwg
On Thu, Sep 17, 2015 at 12:07 PM, Bill Gosper <billgosper@gmail.com> wrote:
on the milliramanujanometer:
No, but In[370]:= N[Sum[n/(E^(2*Sqrt[11]*n*Pi) - 1), {n, 1, Infinity}] == 1/24 - 1/(8*Sqrt[11]*Pi) - ((8 + (8*Sqrt[11])/(3*Sqrt[3] + 7*Sqrt[11])^(1/3) + Sqrt[11]*(3*Sqrt[3] + 7*Sqrt[11])^(1/3))*Gamma[1/22]* Gamma[5/22]*Gamma[3/11]*Gamma[15/22])/(2112*2^(8/11)* Sqrt[11]*Pi^2*Gamma[7/22]*Gamma[10/11]), 69] Out[370]= True might. --rwg
Sum[n/(-1 + E^(2*Sqrt[11]*n*Pi)), {n, 1, Infinity}] == 1/24 - 1/(8*Sqrt[11]*Pi) - ((8 - (4*6^(1/3)*(-6127 + 63*Sqrt[33]))/ (8199 + 56809*Sqrt[33])^(2/3) + (99 + 205*Sqrt[33])/(6*(8199 + 56809*Sqrt[33]))^ (1/3))*Gamma[1/22]*Gamma[5/22]*Gamma[3/11]* Gamma[15/22])/(2112*2^(8/11)*Sqrt[11]*Pi^2* Gamma[7/22]*Gamma[10/11]) --rwg
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Bill Gosper