A340216 - OEIS

%I #35 Dec 22 2024 15:25:03

%S 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,

%T 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,

%U 5,9,7,6,3,2,2,5,6,0,6,9,9,0,6,6,9,0,3

%N Decimal expansion of the series of the reciprocals of the squares of the positive triangular numbers.

%C From _Martin Renner_, Apr 18 2023: (Start)

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

%C 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)

%F 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.

%F 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

%F Equals 4*(2*zeta(2) - 3), with zeta(2) = A013661. - _Wolfdieter Lang_, May 25 2023

%e 1.159472534785811491779321333168201513751599209654387501884465834960...

%p sum(4/(k^2*(k+1)^2),k=1..infinity);

%p sum(1/sum(k^3,k=1..n),n=1..infinity);

%p evalf[100](4/3*Pi^2-12); # _Martin Renner_, Apr 18 2023

%t RealDigits[4*Pi^2/3 - 12, 10, 100][[1]] (* _Amiram Eldar_, Jan 01 2021 *)

%o (PARI) sumpos(n=1, 1/(n*(n+1)/2)^2) \\ _Michel Marcus_, Jan 01 2021

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

%Y Equals 4*A145426.

%K nonn,cons

%O 1,3

%A _Jon E. Schoenfield_, Dec 31 2020