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%I #16 Oct 02 2022 00:01:37
%S 1,8,7,7,9,3,0,8,9,3,6,9,2,8,3,2,7,2,4,4,4,5,7,2,5,8,2,3,5,3,8,0,7,6,
%T 1,5,8,0,5,1,7,5,5,2,3,1,5,5,2,5,5,4,4,7,9,7,4,9,8,3,0,6,5,1,9,4,2,4,
%U 6,7,2,5,8,1,1,0,0,3,2,8,9,4,1,3,8,2
%N Decimal expansion of 1/24 - 1/(8*Pi).
%C Ramanujan's question 387 in the Journal of the Indian Mathematical Society (IV, 120) asked "Show that Sum_{k>=1} k/(exp(2*Pi*k) - 1) = 1/24 - 1/(8*Pi)".
%D G. H. Hardy, P. V. Sheshu Aiyar and B. M. Wilson, Collected Papers of Srinivasa Ramanujan, Cambridge University Press, 1927, p. 326, Q. 427.
%D Oskar Schlömilch, Ueber einige unendliche Reihen, Sitzungsberichte der mathematisch-naturwissenschaftlichen Klasse der Sächsischen Akademie der Wissenschaften, Leipzig, 29 (1877), 101-105.
%H B. C. Berndt, Y. S. Choi and S. Y. Kang, <a href="https://faculty.math.illinois.edu/~berndt/jims.ps">The problems submitted by Ramanujan to the Journal of Indian Math. Soc.</a>, in: Continued fractions, Contemporary Math., Vol. 236 (1999), pp. 15-56 (see Q387, JIMS IV).
%H B. C. Berndt, Y. S. Choi and S. Y. Kang, <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.39.9228">The problems submitted by Ramanujan to the Journal of Indian Math. Soc.</a>, in: Continued fractions, Contemporary Math., Vol. 236 (1999), pp. 15-56 (see Q387, JIMS IV).
%H S. Ramanujan, <a href="https://www.imsc.res.in/~rao/ramanujan/collectedpapers/question/q387.htm">Question 387</a>, Indian Mathematical Society (IV, 120).
%H C. C. Yalavigi, <a href="https://fq.math.ca/Scanned/8-5/advanced8-5.pdf">Problem H-176</a>, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 8, No. 5 (1970), p. 488; <a href="https://www.fq.math.ca/Scanned/10-2/advanced10-2-a.pdf">Keepeing the Q's on Cue</a>, Solution to Problem H-176 by Clyde A. Bridger, ibid., Vol. 10, No. 2 (1972), pp. 186-190.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%e 0.00187793089369283272444572582353807615805175523155255447974983...
%t RealDigits[1/24 - 1/(8*Pi), 10, 100][[1]] (* _Amiram Eldar_, Feb 02 2022 *)
%o (PARI) 1/24 - 1/(8*Pi)
%o (PARI) suminf(k=1,k/(exp(2*Pi*k)-1))
%K nonn,cons
%O -2,2
%A _Hugo Pfoertner_, Sep 24 2018
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