A251992 Decimal expansion of the double infinite sum (negated) sum_{m=1..infinity} sum_{k=0..infinity} (-1)^m/((2k+1)^2+m^2).

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, 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, 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

Offset: 0

Author

Jean-François Alcover, Dec 12 2014

Keywords

• nonn
• cons
• easy

Examples

			-0.480751144424109780520862631352408574248444731679469...
		

Links

• MathOverflow, How to calculate the infinite sum of this double series?

Crossrefs

Cf. A175573.

Programs

Mathematica

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

Formula

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

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

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).