A new form for the MRB constant

Ideas
, ,
1
MRB=\sum_{n=1}^\infty (-1)^n( n^{1 \over n}-1)
=\sum_{m=1}^\infty (-1)^m\bigg( \exp\bigg( {\log(m) \over m}\bigg)-1\bigg)
\begin{array} {rclll} -\exp\bigg( {\log(1) \over 1}\bigg)+1 & = & -{\log(1) \over 1} &-{\log(1)^2 \over 1^2 2!} &-{\log(1)^3 \over 1^3 3!} & - \cdots \\ +\exp\bigg( {\log(2) \over 2}\bigg)-1 & = &+{\log(2) \over 2} &+{\log(2)^2 \over 2^2 2!} &+{\log(2)^3 \over 2^3 3!} & + \cdots \\ -\exp\bigg( {\log(3) \over 3}\bigg)+1 & = &-{\log(3) \over 3} &-{\log(3)^2 \over 3^2 2!} &-{\log(3)^3 \over 3^3 3!} & - \cdots \\ \vdots \qquad & \vdots & \quad\vdots & \quad\vdots& \quad\vdots & \ddots \\ \hline \\ MRB \qquad & = & {\eta^{(1)}(1) \over 1!} &- {\eta^{(2)}(2) \over 2!} &+ {\eta^{(3)}(3) \over 3!} & - \cdots \end{array}.

So,

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I don’t think I should continue to say nul= \sum _{m=2}^{\infty } (-1)^m \left( \eta ^{(m)}(m)\right).

That Is why nul means “No.”

But here is the work with just the given value of nul:

sum = 1/2 Log[2] (-2 EulerGamma + Log[2]);

 nul = -0.07608671642673194446;

  mb = -(sum + nul/E)

0.18785964246206712025

 MRB = 
 NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
  Method -> "AlternatingSigns", WorkingPrecision -> 30]

0.18785964246206712024857897184

 MRB - mb

0.*10^-21

3 Likes
2

10,000 digits of nul = -0.0760867..., the hard to calculate nul, not necessarily \sum _{m=2}^{\infty } (-1)^m \left( \eta ^{(m)}(m)\right)=-0.0724829...?.

m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 10000];
   sum = 1/2 Log[2] (-2 EulerGamma + Log[2]):

nul = (-m - sum)*E

-0.07608671642673194446000…

Here is how to try compute it without computing the MRB constant.
First compute MRBeta2toinf= \sum _{m=2}^{\infty } (-1)^m \frac{\left( \eta ^{(m)}(m)\right)}{m!}.
That is the Cradall first eta formula for MRB -sum.

Print["Start time is " "Start time is ", ds = DateString[], "."];
prec = 10000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{lg, a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 16(*=
    4*number of physical cores*), tsize = 2^7, chunksize, start = 1, 
    ll, ctab, pr = Floor[1.0002 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["Will give ", end, 
    " time estimates, each more accurate than the previous."];
   Print["Will stop at ", end*chunksize, 
    " iterations to ensure precsion of around ", pr, 
    " decimal places."]; d = ChebyshevT[n, 3];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = pr/2^6;
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        lg = Log[ll]/(ll); x = N[E^(lg), iprec];
        pc = iprec;
        While[pc < pr, pc = Min[4 pc, pr];
         x = SetPrecision[x, pc];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pc] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 2 pc] ll (ll - 1)/(3 ll t2 + t^3 z))];
        x - lg, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
       Method -> "EvaluationsPerKernel" -> 16]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], Method -> "EvaluationsPerKernel" -> 16];
    s += ctab.(xvals - 1);
    start += chunksize;
    st = SessionTime[] - T0; kc = k*chunksize;
    ti = (st)/(kc + 10^-10)*(n)/(3600)/(24);
    If[kc > 1, 
     Print[kc, " iterations done in ", N[st - stt, 4], " seconds.", 
      " Should take ", N[ti, 4], " days or ", ti*3600*24, 
      "s, finish ", DatePlus[ds, ti], "."], 
     Print["Denominator computed in  ", stt = st, "s."]];, {k, 0, 
     end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRBeta2toinf = expM[prec];]; Print["Finished on ", 
 DateString[], ". Processor and total time were ", 
 t2[[1]], " and ", st, " s respectively."];

Then

-(MRBeta2toinf*E)

-0.07608671642673194446000…