OFFSET
1,3
COMMENTS
Ramanujan's question 642, part 2, in the Journal of the Indian Mathematical Society (VII, 80) asked: "Show that Sum_{n>=0} (Sum_{k=0..n} 1/(2*k + 1)) * 9^(-n)/(2*n + 1) = Pi^2/8 - (3/8)*log(2)^2".
LINKS
B. C. Berndt, Y. S. Choi, and S. Y. Kang, The problems submitted by Ramanujan to the Journal of Indian Math. Soc., in: Continued fractions, Contemporary Math., 236 (1999), 15-56 (see Q642, JIMS VII).
B. C. Berndt, Y. S. Choi, and S. Y. Kang, The problems submitted by Ramanujan to the Journal of Indian Math. Soc., in: Continued fractions, Contemporary Math., 236 (1999), 15-56 (see Q642, JIMS VII).
EXAMPLE
1.0535306699168442931041479276120195275152550690571442332...
MATHEMATICA
RealDigits[Pi^2/8 - 3*Log[2]^2/8, 10, 120][[1]] (* Amiram Eldar, Jun 27 2023 *)
PROG
(PARI) Pi^2/8-(3/8)*log(2)^2
(PARI) suminf(n=0, sum(k=0, n, 1/(2*k+1))*9^(-n)/(2*n+1))
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Hugo Pfoertner, Sep 19 2018
STATUS
approved