A226985 Sum of inverse of increasing integers with a difference of 0, 1, 2, 3, ...: 1 + 1/2 + 1/4 + 1/7 + 1/11 + 1/16 + 1/22 + 1/29 + 1/37 + ....

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

Offset: 1

Author

Didier Guillet, Jun 25 2013

Keywords

• nonn
• cons

Comments

This is a convergent series since the denominator is quadratic.

We can note that tanh(sqrt(7)*Pi/2) = 0.9995... which is close to 1 by 0.05% so this constant is very close to 2*Pi/sqrt(7). - Didier Guillet, Jul 12 2013

Examples

			2.3736546754401077643216861222374324519138059094067120296733133891251...
		

Crossrefs

Cf. A000124.

Programs

Mathematica

RealDigits[2*Pi*Tanh[Sqrt[7]*Pi/2]/Sqrt[7], 10, 110][[1]] (* Giovanni Resta, Jun 26 2013 *)

PARI

sumpos(k=1,1/(1+k*(k-1)/2)) \\ Charles R Greathouse IV, Jun 26 2013

PARI

2*Pi*tanh(sqrt(7)*Pi/2)/sqrt(7) \\ Charles R Greathouse IV, Jun 26 2013

PARI

sumnumrat(2/(x^2-x+2),1) \\ Charles R Greathouse IV, Feb 04 2025

Formula

Sum_{k >= 1} 1/(1+k*(k-1)/2).

It equals 2*Pi*tanh(sqrt(7)*Pi/2)/sqrt(7). - Giovanni Resta, Jun 26 2013

Extensions

a(12)-a(87) from Giovanni Resta, Jun 26 2013