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%I #11 Feb 07 2019 02:12:35
%S 1,1,8,4,9,0,5,2,0,2,9,4,8,6,4,5,0,2,2,7,3,9,3,5,7,1,2,6,3,8,3,2,0,2,
%T 4,1,2,6,7,8,0,2,2,5,1,7,7,6,4,9,6,9,0,1,5,1,4,1,5,2,2,4,3,5,7,5,1,0,
%U 5,3,0,0,0,7,4,3,7,4,0,3,7,4,0,3,1,0,4
%N Decimal expansion of the limit of (Sum_{j=1..k} arcsin(1/j)) - log(k) as k approaches infinity.
%C Draw a set of circular arcs, with each arc having its center at the origin and drawn counterclockwise so that it connects the line x=1 to the y-axis, and with the j-th arc having radius j; i.e., draw
%C - the 1st arc from (1, 0) to (0, 1),
%C - the 2nd arc from (1, sqrt(3)) to (0, 2),
%C - the 3rd arc from (1, sqrt(8)) to (0, 3),
%C ...
%C - the k-th arc from (1, sqrt(k^2 - 1)) to (0, k).
%C Let alpha(k) be the angle covered by the k-th arc: the first arc is a quarter circle, so alpha(1) = Pi/2; alpha(2) = arcsin(1/2) = 0.523598...; alpha(3) = arcsin(1/3) = 0.339836...; and, in general, alpha(k) = arcsin(1/k), which approaches 1/k as k increases.
%C Let s(k) = Sum_{j=1..k} alpha(j) = Sum_{j=1..k} arcsin(1/j); since arcsin(1/j) > 1/j for j >= 1, and Sum_{j=1..k} 1/j diverges as k increases, s(k) must likewise diverge as k increases.
%C Lim_{k->infinity} ((Sum_{j=1..k} 1/j) - log(k)) approaches the Euler-Mascheroni constant, 0.5772156649015... (cf. A001620); this sequence gives the decimal expansion of lim_{k->infinity} (s(k) - log(k)).
%C (As k increases, the length of the k-th arc approaches 1, and (sum of the lengths of the first k arcs) - k also approaches a constant; see A323737.)
%F Lim_{k->infinity} (Sum_{j=1..k} arcsin(1/j)) - log(k).
%e 1.18490520294864502273935712638320241267802251776496...
%Y Cf. A001620, A323737.
%K nonn,cons
%O 1,3
%A _Jon E. Schoenfield_, Feb 03 2019