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Decimal expansion of Sum_{n>=1} 1/(n^(log(n)^2)) = Sum_{n>=1} exp(-log(n)^3).
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%I #14 Jul 31 2021 00:50:10

%S 2,0,7,1,3,8,4,3,5,3,5,9,8,1,7,8,6,1,8,3,5,9,1,9,8,3,0,7,3,9,1,3,4,7,

%T 2,0,9,4,6,0,9,8,2,4,7,8,2,3,7,4,9,9,6,0,2,9,6,9,1,9,0,5,6,1,9,3,3,4,

%U 1,8,3,5,9,2,7,7,0,1,4,2,8,1,0,8,4,7,6,5,8,0,8,5,8,9,5,4,9,9,9,7,0,9,2,6

%N Decimal expansion of Sum_{n>=1} 1/(n^(log(n)^2)) = Sum_{n>=1} exp(-log(n)^3).

%C An infinite sum that converges faster than A099870.

%C Note that as p > 0 gets larger and larger, the series Sum_{n>=1} 1/(n^(log(n)^p)) converges faster and faster, but will always converge more slowly than Sum_{n>=0} 1/a^n for every a > 1.

%H Jianing Song, <a href="/A346670/b346670.txt">Table of n, a(n) for n = 1..1000</a>

%e 2.07138435359817861835919830739134720946...

%o (PARI) sumpos(n=1, 1/(n^(log(n)^2)))

%Y Cf. A099870, A099871, A073009.

%K nonn,cons

%O 1,1

%A _Jianing Song_, Jul 28 2021