At the Seattle Joint Meeting last week, NeilB delivered gosper.org/mizan.pdf, my brief tribute to the late q-hypergeometric wizard Mizan Rahman, co-author (with George Gasper) of Basic Hypergeometric Series. My note mentions Ramanujan's sum of dyadic rationals converging rapidly to 1/π, Sum[(42 k + 5)*Binomial[2 k, k]^3/2^(12 k + 4), {k, 0, oo}], unwittingly rediscovered in Acceleration of Series - DSpace@MIT <https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwjF5Ka_mqvKAhUW5mMKHf77ALcQFggdMAA&url=http%3A%2F%2Fdspace.mit.edu%2Fbitstream%2Fhandle%2F1721.1%2F6088%2FAIM-304.pdf&usg=AFQjCNHtazNqzTgrSOupXFXaeClJBi3FZg&sig2=yvgV1UOBK4YbSjjWv6ZOdA> . Expecting nothing more than 5/16 HypergeometricPFQ[{47/42, 1/2, 1/2, 1/2}, {1, 1, 5/42}, 1/64] I randomly typed it into Mathematica: Sum[(42 k + 5)*Binomial[2 k, k]^3/2^(12 k + 4), {k, 0, \[Infinity]}] // FunctionExpand 1/\[Pi] + (5 EllipticK[1/16 (8 - 3 Sqrt[7])]^2)/(4 \[Pi]^2) - ( 5 Gamma[1/7]^2 Gamma[2/7]^2 Gamma[4/7]^2)/(64 Sqrt[7] \[Pi]^4) I.e. it got 1/π plus an exotic 0 containing one of the EllipticK s for the 7th singular value! Solving for it, EllipticK[1/16 (8 - 3 Sqrt[7])] == (Gamma[1/7] Gamma[2/7] Gamma[4/7])/(4 7^(1/4) \[Pi]) (an old result). Checking, In[548]:= ContinuedFraction[EllipticK[1/16 (8 - 3 Sqrt[7])]/ EllipticK[1 + 1/16 (-8 + 3 Sqrt[7])], 22] Out[548]= {0, 2, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4} Well, I guess Wolfram isn't just regurgitating canned results! I probably mentioned earlier Mma startling me by evaluating the infinite product of factorials divided by (an improved) Stirling's formula, (d15) in http://www.tweedledum.com/rwg/idents.htm, a result Askey once considered "wild". Now if Mma could only denest radicals. --rwg I recently had occasion to measure the diagonal of an 8" square: 11+1/π inches. Oops, sorry teacher, you want mixed numbers: 11 1/π"