Veit, it's obvious for t=π/2. I was able to squeeze 90+digits of π out of t=1, but some of those Borwein frauds can give you thousands: http://www.math.grinnell.edu/~chamberl/courses/444/worksheets/high-precision... —rwg n 2018-08-29 12:01, Veit Elser wrote:

I suspect it’s true, not fraudulent.

Try:

p = 42;

Simplify[D[ 2 t + 2 Sum[(Cosh[n t] (Cos[t] Sech[t])^n Sin[n t])/n, {n, p}], {t, p}] /. t -> Pi/2]

-Veit

...

On Aug 29, 2018, at 1:14 PM, Bill Gosper <billgosper@gmail.com> wrote:

π == 2t + 2Sum[(Cosh[n*t]*(Cos[t]*Sech[t])^n*Sin[n*t])/n, {n,∞}], (1≤t<2 ?)

Is anybody playing with this? I find it incredible. Is it one of those Borwein or Zagier type high precision frauds? Bibasic telescopy? Am I missing something completely obvious? Inconclusive plotting suggests it holds in a region of the complex plane shaped like a football silhouette centered at t = π/2 + 0i of width > 1 and height < i. (DIY or stay tuned.) —rwg