$C$MRB is approximately 0.1878596424620671202485179340542732. See this and this.
$\sum_{n=1}^\infty (-1)^n\times(n^{1/n}-a)$ is formally convergent only when $a =1$. However, if you extend the meaning of $\sum$ through "summation methods", whereby series that diverge in one sense converge in another sense (e.g. Cesaro, etc.) you get results for other $a$. A few years ago it came to me to ask what value of $a$ gives $$\sum_{n=1}^\infty (-1)^n\times(n^{1/n}-a)=0\text{ ?}$$(For what value of a is the Levin's u-transform's and Cesàro's sum result 0 considering weak convergence?)
The solution I got surprised me: it was $a=1-2\times C\mathrm{MRB}=0.6242807150758657595029641318914535398881938101997224\ldots$. Where $C\mathrm{MRB}$ is $\sum_{n=1}^\infty (-1)^n\times(n^{1/n}-1)$. If that's correct can you explain why?
To see this for yourself in Mathematica enter
FindRoot[NSum[(-1)^n*(n^(1/n) - x), {n, 1, Infinity}], {x, 1}] where 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 wrote Mathematica to find out what method it used and they wrote the following:
Hello Marvin,
Thank you for your email.
If you do not specify a Method for NSum it will try to choose between the EulerMaclaurin or WynnEpsilon methods. In any case, some implicit assumptions about the functions you are summing have to be made. If these assumptions are not correct, you may get inaccurate answers. [http://reference.wolfram.com/mathematica/tutorial/NumericalEvaluationOfSumsAndProducts.html]
From the notes on our internel implementations [http://reference.wolfram.com/mathematica/tutorial/SomeNotesOnInternalImplementation.html]:
Numerical Sums and Products
If the ratio test does not give 1, the Wynn epsilon algorithm is applied to a sequence of partial sums or products.
Otherwise Euler[Dash]Maclaurin summation is used with Integrate or NIntegrate.
[more] 1-2$C$MRB is what I call M2. Here is the formula for M2 by way of Levin-Type Sequence Transformations: in Maple
Digits := 20; fsolve(sum((-1)^j*(j^(1/j)-a), j = 1 .. infinity) = 0, a),
giving 0.62428071507586575950,
and in Mathematica,
Block[{$MaxExtraPrecision = 10000}, FindRoot[NSum[(-1)^j*(j^(1/j) - a), {j, Infinity}], {a, 0.6}, WorkingPrecision -> 20]],
giving {n -> 0.62428071507608096085}.
There is a discrepancy between the two results, but they do agree that M2 is approximately = 0.624280715076.
Finally let a = M2 = $1-2\times C$MRB = 0.6242807150758... and the two limit-points of the series $\sum_{n=1}^\infty (-1)^n\times(n^{1/n}-M2)$ are +/- $C$MRB with its Levin's u-transform's result being 0. See here.
Any explanation of why this is so (or not so) or editing help would be appreciated.