I know of two separate sequences which give approximations to the MRB constant:as partial sums
Let 5^3+50 be just a sample large number
f[x_, n_] := (Log[x]/x)^n/n!; g[x_] := (x^(1/x) - 1)
then
(NSum[(-1)^x (f[x, n]), {x, 1, Infinity}, {n, 1, 5^3 + 50},
Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
and
(NSum[(-1)^x g[x], {x, 1, Infinity}, Method -> "AlternatingSigns",
WorkingPrecision -> 90, NSumTerms -> 100])
both give 100 digits of the constant.
According to what I have found out it is unknown whether the constant is irrational.
However, here is an operation with a twist that gives a seemingly rational result from the two series: In the following, my first suspicion is that a great number of the terms of the two series cancel each other out
Let
a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1,
5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
18.9738238886687791451003113394815962356462125849140174033724730930372\
0038803068158297872*)
and
b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1,
5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
-18.598104603744644904603275471373049775534406395113739831028463724858\
24592490136115559301*)
Then
a/b + 101/99
gives 0. to a great deal of accuracy.
Using another large number we have the dialog that follows:
a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1,
10^3}, Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
18.9738238886687791451003113394815962356462125849140174033724730930372\
0038803068158297872*)
b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1,
10^3}, Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
-18.598104603744644904603275471373049775534406395113739831028463724858\
24592490136115559301*)
a/b
(*\
-1.0202020202020202020202020202020202020202020202020202020202020202020\
20202020202020202020*)
a/b + 101/99
(* 0.*10^-88*)
And changing of the upper limit of x to match the upper limit of n gives the following dialog:
a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, 5^3 + 50}, {n, 1,
5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
17.4647396051636702353546828200037984460733942552094333462726566948090\
9435574173425497138*)
b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, 5^3 + 50}, {n, 1,
5^3 + 50}, Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
-17.118903177338647062377362368122535110507584467977563379019732799862\
37961602407615091254*)
a/b + 101/99
(*0.*10^-88*)
However, the result is probably only close to being rational!!!
Because with a sample smaller number for the upper bound of the summation, u. a/b Is directly related to u.
a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1,
50}, Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
9.58084176556542313267441463676793473285105783990707809477223888856333\
8809797670898335791*)
b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1,
50}, Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
-9.2051224806412888921773787686593882727392516501068005224282295203843\
84346668350470950074*)
a/b + 51/49
(* 0.*10^-89*)
And again changing of the upper limit of x to match the upper limit of n gives the following dialog:
a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, 50}, {n, 1, 50},
Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
11.6400420049405567712416347589817738386513218016649258684668924395814\
104491180570624594618*)
b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, 50}, {n, 1, 50},
Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100])
(*
-11.183569769452691799820394180198174864586564083952575834409367245872\
3355295447999227551691*)
a/b + 51/49
(* -0.*10^-90*)
So we had an upper limit of 50 and a/b + (50+1)/(50-1)==0.
That pattern holds for many small upper limits:
Table[
a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u},
Method -> "AlternatingSigns", WorkingPrecision -> 400,
NSumTerms -> 100]);
b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u},
Method -> "AlternatingSigns", WorkingPrecision -> 400,
NSumTerms -> 100]);
Print[N[a/b + (u + 1)/(u - 1), 30]], {u, 10, 100, 10}]
(* -4.26437037879890888209403349528*10^-14
-3.45296975923093128282587700610*10^-30
-9.13683371485136018997871584145*10^-49
-8.71088418964640386964920759895*10^-69
-7.12931012161710951343301078693*10^-90
-8.25602003178565088281349655952*10^-112
-1.87851117361012541401103121417*10^-134
-1.05980261544468753040510454976*10^-157
-1.76367310964213439705827747685*10^-181
-9.90529404507625197397486146716*10^-206*)
But the 0 is not found by the same pattern for larger upper limits:
Table[a = (NSum[(-1)^x (f[x, n] + g[x]), {x, 1, Infinity}, {n, 1, u},
Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100]);
b = (NSum[(-1)^x (f[x, n] - g[x]), {x, 1, Infinity}, {n, 1, u},
Method -> "AlternatingSigns", WorkingPrecision -> 90,
NSumTerms -> 100]);
Print[a/b + (u + 1)/(u - 1)], {u, 10^3, 10^4, 10^3}]
(* -0.018200018200018200018200018200018200018200018200018200018200018200018200018200018200018
-0.019201519951895139488936387385611997918150994689263823831107472928383383610997417900870
-0.019535131239032539466017291959272619492692850478779121660081980188682422760448769118326
-0.019701895170762387566588616851182492592845180992217751407548856911197496343782915425826
$Aborted*)
I will try to find out why this all happens. You can help me if you wish! (Even if you do I will no doubt make my own contribution too.) When I make these post I always hope that someone will build upon them and perhaps do better research! Like RICHARD J. MATHAR
did with my integral analog of the MRB constant at http://arxiv.org/pdf/0912.3844v3.pdf .
My first suspicion was that a great number of the terms of the two series cancel each other out., bu why then does the operation behave so differently with smaller numbers?