MaplePrimes - comments on Post, MRB Constant F

MaplePrimes - comments on Post, MRB Constant F http://www.mapleprimes.com/posts/35778-MRB-Constant-F en-us 2026 Maplesoft, A Division of Waterloo Maple Inc. Maplesoft Document System Thu, 11 Jun 2026 11:51:45 GMT Thu, 11 Jun 2026 11:51:45 GMT The latest comments added to the Post, MRB Constant F http://www.mapleprimes.com/images/mapleprimeswhite.jpg MaplePrimes - comments on Post, MRB Constant F http://www.mapleprimes.com/posts/35778-MRB-Constant-F evalf,Sum http://www.mapleprimes.com/posts/35778-MRB-Constant-F?ref=Feed:MaplePrimes:MRB Constant F:Comments#comment45092 Have you read the ?evalf,Sum help-page? acer The latest comments added to the Post, MRB Constant F 45092 Mon, 08 Feb 2010 13:07:22 Z acer acer It seems that http://www.mapleprimes.com/posts/35778-MRB-Constant-F?ref=Feed:MaplePrimes:MRB Constant F:Comments#comment45093 Thank you acer. It would sure help if I can figure out what the Levin's u-transform's result given through Maple says about the divergent series. I've only started to look up the meaning of it. If anyone can explain it, that would be great!   Through experimentation it seems that MRB2=a=0.62... where   limit(sum((-1)^j*(j^(1/j)-a), j = 1 .. N-1)-(sum((-1)^j*(j^(1/j)-a), j = 1 .. N)), N = infinity)  = 0.376... approximately = 2* MRB*constant   The latest comments added to the Post, MRB Constant F 45093 Wed, 10 Feb 2010 01:38:15 Z Marvin Ray Burns Marvin Ray Burns summary of f(a) = Sum[(-1)^n*(n^(1/n)-a),{n,Infinity}] for a/=1 http://www.mapleprimes.com/posts/35778-MRB-Constant-F?ref=Feed:MaplePrimes:MRB Constant F:Comments#comment45095 For a not = 1,  series of the form f(a) = Sum[(-1)^n*(n^(1/n)-a),{n,Infinity}] are divergent . I wanted to set the set f(a) = to 0, solve for a and call that value for a "MRB2." I did some bisection method and then resorted to solvers. Using Levin-Type Sequence Transformations Maple and Mathematica were willing to give me a zero, not withstanding, with differing results beyond machine precision even with the use of methods and options. The two results were suspiciously close to MRB2 = 1-1-2M = 0.6242807150758... where M is the MRB constant( that is Sum[(-1)^n*(n^(1/n)-1),{n,Infinity}] ). Now I would like to do two things: Find out precisely where the two limit-points of the series f(a) are converging to when a = MRB2 ( It seems to be +/- MRB constant. ) and find a non trivial proof that shows that MRB2 = 1-2M. The latest comments added to the Post, MRB Constant F 45095 Tue, 16 Feb 2010 23:43:54 Z Marvin Ray Burns Marvin Ray Burns The two limit-points=+/- MRB constant http://www.mapleprimes.com/posts/35778-MRB-Constant-F?ref=Feed:MaplePrimes:MRB Constant F:Comments#comment45096 Let a = MRB2 = 1-1-2M = 0.6242807150758..., where M is the MRB constant and the two limit-points of the series f(a) = Sum[(-1)^j*(j^(1/j) - a), {j, Infinity}] converge to +/- MRB constant with its Levin's u-transform's result being 0. The latest comments added to the Post, MRB Constant F 45096 Tue, 16 Feb 2010 23:45:14 Z Marvin Ray Burns Marvin Ray Burns Maple shows that a=1-2M http://www.mapleprimes.com/posts/35778-MRB-Constant-F?ref=Feed:MaplePrimes:MRB Constant F:Comments#comment45097 Since M = the MRB constant = sum((-1)^m*(m^(1/m)-1), m = 1 .. infinity) Maple shows that sum((-1)^m*(m^(1/m)-a), m = 1 .. infinity) = 0 for a = MRB2 = 1- 2*MRB constant. The Maple input, evalf(sum((-1)^n*(n^(1/n)-1+2*evalf(sum((-1)^m*(m^(1/m)-1), m = 1 .. infinity), 20)), n = 1 .. infinity), 20); gives approximately 0 and the larger the precision, (Here it is 20 digits.), the closer the sum is to zero. When the precision is infinate, zero minus the sum, (Which sum does exist.), equals zero. Thus the sum is, at that time, zero. So I have achieved my two goals which were: find out what value the two limit points of sum((-1)^m*(m^(1/m)-a), m = 1 .. infinity) for a=MRB2 were and prove that MRB2 = 1-2M. <img alt="" src="file:///C:/Users/Owner/AppData/Local/Temp/moz-screenshot-12.png" /> The latest comments added to the Post, MRB Constant F 45097 Tue, 16 Feb 2010 23:55:44 Z Marvin Ray Burns Marvin Ray Burns MRB2=1-2M http://www.mapleprimes.com/posts/35778-MRB-Constant-F?ref=Feed:MaplePrimes:MRB Constant F:Comments#comment45094 Here is the formula for MRB2 in Maple: <img alt="" src="file:///C:/Users/Owner/AppData/Local/Temp/moz-screenshot-7.png" /> Digits := 50; fsolve(sum((-1)^j*(j^(1/j)-n), j = 1 .. infinity) = 0, n), giving 0.6242807150758657595029641318914535398881938101997; and in Mathematica: Block[{$MaxExtraPrecision = 10000},  FindRoot[NSum[(-1)^j*(j^(1/j) - n), {j, Infinity}], {n, 0.6},   WorkingPrecision -> 50]], giving {n -> 0.62428071507608096085825341581276859415083890780807}. There is a huge discrepancy between the two results. But they agree that MRB2 approximately = 0.624280715076. However, according to Richard Mathar's trivial solution, let M=the MRB constant and MRB2=1-2M=0.62428071507586575950296413189145353988819381019972242765599 063182104553687067957259340669113378500619231530828748396187753725981048 15467391221413022076317583253267478367727945237472412 ....   The latest comments added to the Post, MRB Constant F 45094 Mon, 08 Mar 2010 03:46:09 Z Marvin Ray Burns Marvin Ray Burns