%I #25 Jan 03 2024 08:44:59
%S 5,1,7,1,0,0,3,7,9,0,4,2,4,0,1,7,2,5,0,6,4,8,1,0,7,2,1,3,1,3,5,7,4,5,
%T 0,4,7,2,5,0,7,3,7,9,0,8,0,6,6,9,2,7,6,5,7,5,6,7,2,5,9,1,5,7,8,7,1,2,
%U 1,1,4,9,2,6,6,7,7,6,2,7,0,1,5,7,8,3,9,1,2,3,1,7,7,8,6,1,5,0
%N Decimal expansion of (1/1) - (1/2+1/3) + (1/4+1/5+1/6) - (1/7+1/8+1/9+1/10) + (1/11+1/12+1/13+1/14+1/15) - ...
%C There are n terms in the n-th group v(n), from 1 / ((n^2-n+2)/2) up to 1 / ((n^2+n)/2).
%C As |v(n+1)| < |v(n)|, this series is convergent according to the alternating series test.
%D Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année, MP, Dunod, 1997, Exercice 3.3.19 pp. 285 and 303 .
%H user115748, <a href="https://mathoverflow.net/q/461393">Power series x^n/n with one plus, then two minuses, then three plusses, and so on</a>, MathOverflow, Jan 2024.
%F Equals Sum_{n>=1} Sum_{k = (n^2-n+2)/2..(n^2+n)/2} (-1)^(n+1) / k.
%F Equals Sum_{n>=1} (-1)^(n+1) * (A081971(n)/A082681(n)).
%e 0.517100379042401725064810772131357...
%p evalf(sum(sum((-1)^(n+1)/k, k= (n^2-n+2)/2..(n^2+n)/2), n=1..infinity),100);
%o (PARI) sumalt(n=1, (-1)^(n+1)*sum(k=(n^2-n+2)/2, (n^2+n)/2, 1/k)) \\ _Michel Marcus_, May 09 2022
%Y Cf. A081971, A082681.
%Y Cf. A002162, A339799 (other harmonic series with + and -).
%K nonn,cons
%O 0,1
%A _Bernard Schott_, May 09 2022