A251992 - OEIS

A251992

Decimal expansion of the double infinite sum (negated) sum_{m=1..infinity} sum_{k=0..infinity} (-1)^m/((2k+1)^2+m^2).

%I #12 Dec 15 2014 09:12:32

%S 4,8,0,7,5,1,1,4,4,4,2,4,1,0,9,7,8,0,5,2,0,8,6,2,6,3,1,3,5,2,4,0,8,5,

%T 7,4,2,4,8,4,4,4,7,3,1,6,7,9,4,6,9,0,2,0,7,5,5,4,7,2,1,3,2,6,8,9,1,0,

%U 8,5,1,7,0,7,7,6,5,3,9,5,3,5,3,5,7,1,5,2,7,7,6,3,1,7,4,3,0,7,5,4,2,2,9

%N Decimal expansion of the double infinite sum (negated) sum_{m=1..infinity} sum_{k=0..infinity} (-1)^m/((2k+1)^2+m^2).

%H MathOverflow, <a href="http://mathoverflow.net/questions/189199">How to calculate the infinite sum of this double series?</a>

%F -Pi*(Pi-log(2))/16.

%F Also equals sum_{m=1..infinity} (-1)^m*Pi*tanh(m*Pi/2)/(4*m).

%F Also equals -Pi^2/16 - (Pi/4)*log(theta_2(0,exp(-Pi))) + (Pi/4)*log(theta_3(0,exp(-Pi))), where 'theta' is the elliptic theta function, that is -Pi^2/16 - (Pi/4)*log(A248557) + (Pi/4)*log(A175573).

%e -0.480751144424109780520862631352408574248444731679469...

%t RealDigits[-Pi*(Pi-Log[2])/16, 10, 103] // First

%Y Cf. A175573.

%K nonn,cons,easy

%O 0,1

%A _Jean-François Alcover_, Dec 12 2014