A327773 - OEIS

%I #13 Feb 16 2025 08:33:58

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

%T 8,9,9,2,8,4,9,9,9,2,7,9,7,3,4,2,2,5,7,0,0,7,7,1,1,7,0,1,8,2,8,8,3,9,

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

%N Decimal expansion of Sum_{k>=1} 1/(k*(k+1))^4.

%C Sum_{k>=1} 1/(k*(k+1)) = 1

%C Sum_{k>=1} 1/(k*(k+1))^2 = -3 + Pi^2/3

%C Sum_{k>=1} 1/(k*(k+1))^3 = 10 - Pi^2

%C Sum_{k>=1} 1/(k*(k+1))^4 = -35 + 10*Pi^2/3 + Pi^4/45

%C Sum_{k>=1} 1/(k*(k+1))^5 = 126 - 35*Pi^2/3 - Pi^4/9

%C Sum_{k>=1} 1/(k*(k+1))^6 = -462 + 42*Pi^2 + 7*Pi^4/15 + 2*Pi^6/945

%C Sum_{k>=1} 1/(k*(k+1))^7 = 1716 - 154*Pi^2 - 28*Pi^4/15 - 2*Pi^6/135

%C Sum_{k>=1} 1/(k*(k+1))^8 = -6435 + 572*Pi^2 + 22*Pi^4/3 + 8*Pi^6/105 + Pi^8/4725

%C Sum_{k>=1} 1/(k*(k+1))^9 = 24310 - 2145*Pi^2 - 143*Pi^4/5 - 22*Pi^6/63 - Pi^8/525

%C Sum_{k>=1} 1/(k*(k+1))^10 = -92378 + 24310*Pi^2/3 + 1001*Pi^4/9 + 286*Pi^6/189 + 11*Pi^8/945 + 2*Pi^10/93555

%C In general, for s > 1, Sum_{k>=1} 1/(k*(k+1))^s = (-1)^s * (-binomial(2*s-1, s-1) + Sum_{j=1..floor(s/2)} ((-1)^(j+1) * binomial(2*s-2*j-1, s-1) * Bernoulli(2*j) * (2*Pi)^(2*j) / (2*j)!).

%C Equivalently, for s > 1, Sum_{k>=1} 1/(k*(k+1))^s = (-1)^s * (-binomial(2*s-1, s-1) + 2*Sum_{j=1..floor(s/2)} (binomial(2*s-2*j-1, s-1) * zeta(2*j)).

%H R. J. Mathar, <a href="https://arxiv.org/abs/1301/6293">Tighly circumscribed regular polygons</a>, arXiv:1301.6293 (2013), see Appendix.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BernoulliNumber.html">Bernoulli Number</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RiemannZetaFunction.html">Riemann zeta function</a>

%F Equals Pi^4/45 + 10*Pi^2/3 - 35.

%e 0.06332780438680511248031072600283958992849992797342257007711701828839...

%p evalf(sum(1/(k*(k+1))^4, k=1..infinity), 120);

%t RealDigits[N[Sum[1/(k*(k + 1))^4, {k, 1, Infinity}], 105]][[1]]

%o (PARI) suminf(k=1, 1/(k*(k+1))^4)

%Y Cf. A145426, A248619.

%K nonn,cons,changed

%O 0,2

%A _Vaclav Kotesovec_, Sep 25 2019