I could use some help in clarifying the following relationship, I've known for a few years.

Let m be the MRB constant

m = 0.187859642462067120248517934054273230055903094900138786172004684089`65.\ 2738334914441;

Below regularization is used so that the sum that formally diverges returns a result that can be interpreted as evaluation of the analytic extension of the series:, I checked a few other methods and got the same results.

Perform the following iteration.

x = m; For[a = 1, a < 10, Print[x = NSum[(-1)^n (n^(1/n) - x), {n, 1, [Infinity]}, WorkingPrecision -> 65, Method -> "AlternatingSigns"]], a++]

gives -1-2^-a (-1+m)+2 m

a =.

and

u = Limit[-1 - 2^-a (-1 + m) + 2 m , a -> [Infinity]]

-0.62428071507586575950296413189145353988819381019972242765599063182

Which is u = 2 m - 1.

Then of course,

u - NSum[(-1)^n (n^(1/n) - u), {n, 1, [Infinity]}, WorkingPrecision -> 65, Method -> "AlternatingSigns"]

gives 0. Where 65 is just max precision.

And perhaps surprisingly,

NSum[(-1)^n (n^(1/n) + u), {n, 1, [Infinity]}, WorkingPrecision -> 65, Method -> "AlternatingSigns"]

gives 0. Where 65 is just max precision.

And both of the following follows.

m + NSum[(-1)^n (n^(1/n) + u), {n, 1, 10^50 + 1}, WorkingPrecision -> 65, Method -> "AlternatingSigns"]

gives 0. Where 10^50+1 is just a big odd number. ie. The Liminf of (-1)^n (n^(1/n)+u) is -m.

m - NSum[(-1)^n (n^(1/n) + u), {n, 1, 10^50}, WorkingPrecision -> 65, Method -> "AlternatingSigns"]

gives 0. Where 10^50 is just a big even number. ie. The Limsup of (-1)^n (n^(1/n)+u) is m.n