A340216 Decimal expansion of the series of the reciprocals of the squares of the positive triangular numbers.

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

Offset

1,3

Comments

From Martin Renner, Apr 18 2023: (Start)

Also decimal expansion of the sum of the reciprocals of the sum of the first n cubes when n approaches infinity.

The partial sums of the series lead to Sum_{k=1..n} 4/(k^2*(k+1)^2) = 8*(Sum_{k=1..n} 1/k^2) - 4*n*(3*n+4)/(n+1)^2 and contains for n approaching infinity the Basel problem solved by Euler in 1734, namely Sum_{k>=1} 1/k^2 = Pi^2/6. (End)

Links

Table of n, a(n) for n=1..87.

Formula

Sum_{k>=1} 1/(k*(k+1)/2)^2 = Sum_{k>=1} 1/A000537(k) = 1/1^2 + 1/3^2 + 1/6^2 + 1/10^2 + ... = 4*Pi^2/3 - 12.

Sum_{n>=1} (1/Sum_{k=1..n} k^3) = 1 + 1/(1^3 + 2^3) + 1/(1^3 + 2^3 + 3^3) + ... = 4*Pi^2/3 - 12. - Martin Renner, Apr 18 2023

Equals 4*(2*zeta(2) - 3), with zeta(2) = A013661. - Wolfdieter Lang, May 25 2023

Example

1.159472534785811491779321333168201513751599209654387501884465834960...

Maple

sum(4/(k^2*(k+1)^2), k=1..infinity);
sum(1/sum(k^3, k=1..n), n=1..infinity);
evalf[100](4/3*Pi^2-12); # Martin Renner, Apr 18 2023

Mathematica

RealDigits[4*Pi^2/3 - 12, 10, 100][[1]] (* Amiram Eldar, Jan 01 2021 *)

Prog

(PARI) sumpos(n=1, 1/(n*(n+1)/2)^2) \\ Michel Marcus, Jan 01 2021

Crossrefs

Cf. A000217 (triangular numbers), A000537, A013661.

Equals 4*A145426.

Sequence in context: A010774 A272610 A350760 * A198748 A196754 A198217

Adjacent sequences: A340213 A340214 A340215 * A340217 A340218 A340219

Keyword

nonn,cons,changed

Author

Jon E. Schoenfield, Dec 31 2020

Status

approved