Dear Marvin,
thank you for your nice answer! I need to add the reference, where I found the Shanks transformation:
Carl M. Bender, Steven A. Orszag; Advanced Mathematical Methods for Scientists and Engineers; Springer, 1999
An outline of the idea goes like this: Suppose the sequence of the partial sums
$A_n$ behaves like:
$$A_n = A + \alpha q^n$$
This actually describes roughly the situation in case of the MRB. (If it were an exact description, we had the "exact" solution after just one single Shanks.) The wanted value of course is
$A$, and because we have a three parameter model, we need three terms of the sequence to calculate
$A$, e.g.
$A_{n-1},A_{n},A_{n+1}$. Then one gets:
$$A = \frac{A_{n+1}A_{n-1}-A_{n}^2}{A_{n+1}+A_{n-1}-2A_{n}}$$
I find the possible improvement by this simple transformation most impressive. It can be illustrated like so (using the definition of sac
from my first post):
And - at last - to answer your question: I do not think that Mathematica has an issue with more Shanks iterations. But then of course one needs to use more terms and more digits:
ClearAll["Global`*"]
m = Table[N[(-1)^k (k^(1/k) - 1), 2000], {k, 1, 2000}];
(* partial sums of the series *)
am = Accumulate@m;
shanks[ac_List] := Table[(ac[[n + 1]] ac[[n - 1]] - ac[[n]]^2)/(ac[[n + 1]] + ac[[n - 1]] - 2 ac[[n]]), {n, 2, Length[ac] - 1}]
sac = NestList[shanks, am, 24];
ListLogPlot[Abs[Differences /@ sac], Joined -> True, GridLines -> Automatic, ImageSize -> Large]
giving (after two or three seconds on my old PC):
Then for an approximation of your MRB constant we get:
err = N[Abs[sac[[-1, -1]] - sac[[-1, -2]]], 5]
(* Out: 6.2079*^-146 *)
numberOfDigits = Floor[-Log[10, err]]
(* Out: 145 *)
mrb = N[sac[[-1, -1]], numberOfDigits]
(* Out: \
0.18785964246206712024851793405427323005590309490013878617200468408947\
7231564660213703296654433107496903842345856258019061231370094759226630\
43892937955325691395341832`145. *)
Regards -- Henrik