OFFSET
1,2
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).
LINKS
Mathematica Stack Exchange, Converges or diverges?, 2021.
EXAMPLE
1.40710442743517658735368769650782855052127407144777551479405092825455...
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 *)
PROG
(PARI) default(realprecision, 120); sumpos(k=1, (1 - log(k)/k)^(2*k))
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Vaclav Kotesovec, May 30 2022
STATUS
approved