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:
m = NSum[(-1)^k (k^(1/k) - 1), {k, 1, Infinity},
WorkingPrecision -> 1000, Method -> "AlternatingSigns"];
N::meprec: Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating -E^(2 I Interval[{0,\[Pi]}])+E^(2 I Interval[{0,\[Pi]}]). >>
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"])]
.
You find there is a pattern to g[n]
For[n = 2, n < 20,
Print[RootApproximant[
MantissaExponent[g[10 n]/g[10 (n + 1)]][[1]]]]; n++]
8/27
27/64
64/125
125/216
216/343
343/512
512/729
729/1000
1000/1331
1331/1728
1728/2197
2197/2744
2744/3375
3375/4096
4096/4913
4913/5832
5832/6859
6859/8000
, and find g[10 n]/g[10 (n + 1)] ->n^3/(n+1)^3 *10 to a factor:
For[n = 2, n < 20, Print[
RootApproximant[
MantissaExponent[
g[10 n]/g[10 (n + 1)]
][[1]]
] - n^3/(n + 1)^3
]; n++]
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0