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%I #12 Jul 22 2020 13:19:58
%S 9,1,5,2,4,3,8,6,0,8,5,6,2,2,5,9,5,9,6,3,4,0,0,9,7,1,9,4,8,4,4,0,8,3,
%T 1,1,8,7,9,0,5,3,9,7,4,0,0,8,0,6,7,0,0,2,1,8,3,2,0,7,9,7,3,3,9,2,7,3,
%U 0,6,1,2,0,9,8,1,7,7,5,8,0,0,5,6,0,7,3,2,4,3,8,9,5,5,2,4,9,1,3,9,5
%N Decimal expansion of 1/(2*tan(1/2)).
%C The divergent series Sum_{n>=1} sin(n) sums to this value if interpreted as a geometric series; that is, Sum_{n>=1} sin(n) = Im(Sum_{n>=1} e^(ni)) = Im(-e^i/(e^i-1)) = 1/(2*tan(1/2)). If x_m = Sum_{0<n<m} sin(n), then (max(x_1, x_2, ...) + min(x_1, x_2, ...))/2 tends to this number. Given by the series 1+Sum_{n>=1} (-1)^n B_(2n)/(2n)!. Corresponding value for cos is -1/2.
%F Equals Sum_{k>=0} (-1)^k * bernoulli(2*k)/(2*k)! = Sum_{k>=0} (-1)^k * A027641(2*k)/(A027642(2*k)*(2*k)!). - _Amiram Eldar_, Jul 22 2020
%e 0.915243860856225...
%t RealDigits[N[1/(2 Tan[1/2]), 101]]
%o (PARI) 1/(2*tan(1/2)) \\ _Michel Marcus_, Jul 22 2020
%Y Cf. A027641, A027642.
%K cons,nonn
%O 0,1
%A _Fredrik Johansson_, Aug 20 2006
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