# A new form for the MRB constant

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 $$MRB=\sum_{n=1}^\infty (-1)^n( n^{1 \over n}-1)$$ $$=\sum_{m=1}^\infty (-1)^m\bigg( \exp\bigg( {\log(m) \over m}\bigg)-1\bigg)$$ $$\begin{array} {rclll} -\exp\bigg( {\log(1) \over 1}\bigg)+1 & = & -{\log(1) \over 1} &-{\log(1)^2 \over 1^2 2!} &-{\log(1)^3 \over 1^3 3!} & - \cdots \\ +\exp\bigg( {\log(2) \over 2}\bigg)-1 & = &+{\log(2) \over 2} &+{\log(2)^2 \over 2^2 2!} &+{\log(2)^3 \over 2^3 3!} & + \cdots \\ -\exp\bigg( {\log(3) \over 3}\bigg)+1 & = &-{\log(3) \over 3} &-{\log(3)^2 \over 3^2 2!} &-{\log(3)^3 \over 3^3 3!} & - \cdots \\ \vdots \qquad & \vdots & \quad\vdots & \quad\vdots& \quad\vdots & \ddots \\ \hline \\ MRB \qquad & = & {\eta^{(1)}(1) \over 1!} &- {\eta^{(2)}(2) \over 2!} &+ {\eta^{(3)}(3) \over 3!} & - \cdots \end{array}.$$So, 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 EulerGamma + Log); 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 Answer
 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 EulerGamma + Log): 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[], " and ", st, " s respectively."]; Then -(MRBeta2toinf*E)  -0.07608671642673194446000... Answer