OFFSET
1,4
COMMENTS
Ramanujan's question 642, part 1, in the Journal of the Indian Mathematical Society (VII, 80) asked: "Show that Sum_{n>=0} (Sum_{k=0..n} 1/(2*k + 1)) * 5^(-n)/(2*n + 1) = Pi^2 / (4*sqrt(5))".
LINKS
B. C. Berndt, Y. S. Choi, 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, 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
1.1034553175933452655171924621290062932191351122702294749...
MATHEMATICA
RealDigits[Pi^2/(4Sqrt[5]), 10, 120][[1]] (* Harvey P. Dale, Feb 21 2020 *)
PROG
(PARI) Pi^2/(4*sqrt(5))
(PARI) suminf(n=0, sum(k=0, n, 1/(2*k+1))*5^(-n)/(2*n+1))
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Hugo Pfoertner, Sep 19 2018
STATUS
approved