I should note that this does not depend on starting with m.

That is very interesting! Input like that was what I posted the question for.

Also I need to redefine a few things so as to provide this post with workable code, since your "code" is in an image rather than actual cut-and-paste form. (Useful rule: provide explicit code in the post when at all possible.)

I failed to mention that all of the code was uploaded an A173273.nb. I didn't post a sample of it because it looks ugly when posted as a code sample:

Let m be the MRB constant=\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(\[Infinity]\)]\(
\*SuperscriptBox[\((\(-1\))\), \(n\)]\ \(\((\(-1\) + 
\*SuperscriptBox[\(n\), 
FractionBox[\(1\), \(n\)]])\)\(.\)\)\)\)

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:

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.