OFFSET
0,1
COMMENTS
Ramanujan's question 606 in the Journal of the Indian Mathematical Society (VI, 239) asked "Show that Sum_{n>=0} (sqrt(5) - 2)^(2*n + 1) / (2*n + 1)^2 = Pi^2/24 - (1/12) * (log(2 + sqrt(5)))^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 Q606, JIMS VI).
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 Q606, JIMS VI).
EXAMPLE
0.2375599012791608147454069882378567272924327127647254563224136358251...
MATHEMATICA
RealDigits[Pi^2/24 - Log[2+Sqrt[5]]^2/12, 10, 120][[1]] (* Amiram Eldar, Jun 27 2023 *)
PROG
(PARI) Pi^2/24-(1/12)*log(2+sqrt(5))^2
(PARI) suminf(k=0, (sqrt(5)-2)^(2*k+1)/(2*k+1)^2)
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Hugo Pfoertner, Sep 16 2018
STATUS
approved