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

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, 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, 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

Offset: 1

Author

Vaclav Kotesovec, May 30 2022

Keywords

• nonn
• cons

Comments

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).

Examples

			1.40710442743517658735368769650782855052127407144777551479405092825455...
		

Links

• Mathematica Stack Exchange, Converges or diverges?, 2021.

Crossrefs

Cf. A091812, A354592, A354593.

Programs

Maple

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:

Mathematica

NSum[(1 - Log[n]/n)^(2*n), {n, 1, Infinity}, WorkingPrecision -> 100, NSumTerms -> 1000] (* only 74 digits are correct *)

PARI

default(realprecision, 120); sumpos(k=1, (1 - log(k)/k)^(2*k))