Again. let g[n] be the error in using f, as you use 10^n terms, as shown in the original post; using the following piece of code, and then it appears to be the case that the g[10 n]'s are rational factors of Log[10]^3, to better and better precision as n gets large:
m =
NSum[(-1)^k (k^(1/k) - 1), {k, 1, Infinity},
WorkingPrecision -> 1000, Method -> "AlternatingSigns"];
f[k_] := Log[(k - Log[k])/k];
ClearAll[g];
g[n_] := Module[{},
m - (NSum[(-1)^k (k^(1/k) - 1 + f[k]), {k, 1, 10^n},
WorkingPrecision -> Floor[3 n + 100],
Method -> "AlternatingSigns"] -
NSum[(-1)^k*f[k], {k, 1, Infinity},
WorkingPrecision -> Floor[3 n + 100],
Method -> "AlternatingSigns"])]
Table[
RootApproximant[MantissaExponent[Log[10]^3/g[10 n]][[1]]], {n, 2, 20}]
(*
{3/20, 4/9, 3/16, 24/25, 5/9, 120/343, 15/64, 40/243, 3/25, \
1200/1331, 25/36, 1200/2197, 150/343, 16/45, 75/256, 1200/4913, \
50/243, 1200/6859, 3/20}
*)
