OFFSET
0,1
COMMENTS
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).
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.
REFERENCES
J. Guégand and M.-A. Maingueneau, Exercices d'Analyse, Exercice 1.18 p. 23, 1988, Classes Préparatoires aux Grandes Ecoles, Ellipses.
LINKS
Mathematics Stack Exchange, Infinite series sum_{n=2..infinity} 1/(n*log(n)).
FORMULA
Limit_{n->oo} 1/(2*log(2) + 1/(3*log(3)) + ... + 1/(n*log(n)) - log(log(n)).
EXAMPLE
0.79467864545289940220389796...
MAPLE
limit(sum(1/(k*log(k)), k=2..n) - log(log(n)), n = infinity);
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Bernard Schott, Apr 08 2023
STATUS
approved