On 2016-03-14 01:02, Jon Ziegler wrote:

This is fun: http://mathwithbaddrawings.com/2016/03/14/the-pi-day-recipe-book A nice collection of ways to make pi.

Regards, Jon

1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 - ... == π/√8 seems to have a nice generalization: Sum[(-1)^Ceiling[n/k]/(1 - 2n), {n, Infinity}] == π Sum[Tan[((2j - 1)*π)/(4k)], {j,1,k}]/(4k), e.g., for k = 9, Sum[(-1)^Ceiling[n/9]/(1 - 2*n), {n, Infinity}] == 1/36 \[Pi] (5 + Cot[\[Pi]/36] + Cot[(5 \[Pi])/36] + Cot[(7 \[Pi])/36] + Tan[\[Pi]/36] + Tan[(5 \[Pi])/36] + Tan[(7 \[Pi])/36]) which shouldn't be hard to prove. I have formulæ for sums and prods of trigs over periods. This tan sum is over a half-period, but with "natural" endpoints. Maybe doable. --rwg