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%I #46 Apr 23 2023 23:52:42
%S 7,9,4,6,7,8,6,4,5,4,5,2,8,9,9,4,0,2,2,0,3,8,9,7,9,6,2,0,6,5,1,4,9,5,
%T 1,4,0,6,4,9,9,9,5,9,0,8,8,2,8,0,4,9,6,8,9,0,1,5,1,2,0,9,5,0,1,4,8,1,
%U 7,8,5,8,9,6,0,6,8,7,5,6,6,6,9,6,6,1,4,7,7,7,6,2,7,3,3
%N Decimal expansion of lim_{n->oo} ( Sum_{k=2..n} 1/(k*log(k)) - log(log(n)) ).
%C Let u(n) = Sum_{k=2..n} 1/(k*log(k)) - log(log(n)), then (u(n)) is strictly decreasing and lower bounded by -log(log(2)) = A074785, so (u(n)) is convergent, while the series v(n) = Sum_{k=2..n} 1/(k*log(k)) diverges (see Mathematics Stack Exchange link).
%C Compare with w(n) = Sum_{k=1..n} 1/k - log(n) that converges (A001620), while the harmonic series H(n) = Sum_{k=1..n} 1/k diverges.
%D J. Guégand and M.-A. Maingueneau, Exercices d'Analyse, Exercice 1.18 p. 23, 1988, Classes Préparatoires aux Grandes Ecoles, Ellipses.
%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/574503/infinite-series-sum-n-2-infty-frac1n-log-n">Infinite series sum_{n=2..infinity} 1/(n*log(n))</a>.
%F Limit_{n->oo} 1/(2*log(2) + 1/(3*log(3)) + ... + 1/(n*log(n)) - log(log(n)).
%F Equals A241005 - log(log(2)) = A241005 + A074785. - _Amiram Eldar_, Apr 08 2023
%e 0.79467864545289940220389796...
%p limit(sum(1/(k*log(k)), k=2..n) - log(log(n)), n = infinity);
%Y Cf. A001620, A074785, A241005.
%K nonn,cons
%O 0,1
%A _Bernard Schott_, Apr 08 2023
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