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)
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
KEYWORD
nonn,cons
AUTHOR
Jon E. Schoenfield, Dec 31 2020
STATUS
approved