OFFSET
0,1
COMMENTS
For p>1, q>1 in R, Sum_{m >1, n>1} 1/(m^p*n^q-1) = Sum_{k>0} (zeta(k*p) - 1) * (zeta(k*q) - 1) [Proof in References]. This sequence corresponds to p = q = 2.
Double inequality: Sum_{m>1, n>1} 1/(m^2*n^2+1) = A338107 = 0.409... < Sum_{m>1, n>1} 1/(m^2*n^2) = (zeta(2)-1)^2 = 0.415... < Sum_{m>1, n>1} 1/(m^2*n^2-1) = this constant = 0.423...
REFERENCES
Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année MP, Dunod, 1997, Exercice 3.25, p. 277.
FORMULA
Equals Sum_{k>0} (zeta(2*k) - 1)^2.
Equals -3/4 + Sum_{k>=2} (1/2 - Pi*cot(Pi/k)/(2*k)). - Vaclav Kotesovec, Oct 14 2020
EXAMPLE
0.4230355257613131597420971016391038628995464... (with help of Amiram Eldar).
MATHEMATICA
RealDigits[Sum[(Zeta[2*k] - 1)^2, {k, 1, 100}], 10, 90][[1]] (* Amiram Eldar, Oct 10 2020 *)
PROG
(PARI) sumpos(k=1, (zeta(2*k) - 1)^2) \\ Michel Marcus, Oct 10 2020
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Bernard Schott, Oct 10 2020
STATUS
approved