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A354450 Decimal expansion of Sum_{k>=1} (1 - log(k)/k)^(2*k).

%I A354450 #26 Aug 12 2022 09:24:23
%S A354450 1,4,0,7,1,0,4,4,2,7,4,3,5,1,7,6,5,8,7,3,5,3,6,8,7,6,9,6,5,0,7,8,2,8,
%T A354450 5,5,0,5,2,1,2,7,4,0,7,1,4,4,7,7,7,5,5,1,4,7,9,4,0,5,0,9,2,8,2,5,4,5,
%U A354450 5,0,1,3,6,4,2,9,0,6,0,8,1,5,2,6,2,8,8,6,5,6,5,1,6,2,8,6,0,0,2,8,8,9,7,9,4
%N A354450 Decimal expansion of Sum_{k>=1} (1 - log(k)/k)^(2*k).
%C A354450 Convergence of this sum is slow, but (1 - log(k)/k)^(2*k) can be expanded as 1/k^2 - log(k)^2/k^3 + (-4*log(k)^3 + 3*log(k)^4)/(6*k^4) + (-3*log(k)^4 + 4*log(k)^5 - log(k)^6)/(6*k^5) + ... and each of these summands can be evaluated exactly because Sum_{k>=1} log(k)^r/k^s is equal to the r-th derivative of zeta(s) * (-1)^r. For example, Sum_{k>=1} log(k)^2/k^3 = zeta''(3).
%H A354450 Mathematica Stack Exchange, <a href="https://mathematica.stackexchange.com/questions/257602/converges-or-diverges">Converges or diverges?</a>, 2021.
%e A354450 1.40710442743517658735368769650782855052127407144777551479405092825455...
%p A354450 Digits := 120: ser := sort(convert(series((1-log(n)/n)^(2*n), n = infinity, 300), polynom), n): s := evalf(sum(op(1, ser), n = 1..infinity) + sum(op(2, ser), n = 1..infinity), 120): for k from 3 to nops(ser) do serx := expand(op(k, ser)): for j to nops(serx) do s := s + evalf(sum(op(j, serx), n = 1..infinity), 120) end do: print(k, s) end do:
%t A354450 NSum[(1 - Log[n]/n)^(2*n), {n, 1, Infinity}, WorkingPrecision -> 100, NSumTerms -> 1000] (* only 74 digits are correct *)
%o A354450 (PARI) default(realprecision, 120); sumpos(k=1, (1 - log(k)/k)^(2*k))
%Y A354450 Cf. A091812, A354592, A354593.
%K A354450 nonn,cons
%O A354450 1,2
%A A354450 _Vaclav Kotesovec_, May 30 2022