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In the question MRB constant proofs wanted , I gave the following excerpt from http://marvinrayburns.com/UniversalTOC25.pdf . part1

part1. I accurately worked formula 44 in Matheamatica in the following code.

(*define the eta function*)

eta[s_] := (1 - 2^(1 - s)) Zeta[s];

(*define the higher derivatives of the eta(0)*)

a[i_] := Derivative[i][eta][0];

(*Define c:*)

c[j_] := Sum[Binomial[j, d](-1)^dd^(j - d), {d, 1, j}]

(*formula (44)*)

N[Sum[c[m]/m!*a[m], {m, 1, 40}], 100]

It gave -0.1878587... .

Can anyone come up with a more lucrative program for formula 44 in Mathematica or Maple?

Glorfindel
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Marvin Ray Burns
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2 Answers2

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N[Sum[c[m]/m!*a[m], {m, 1, 40}], 100] is an overkill on working precision and too finite to give much accuracy. N[Sum[c[m]/m!*a[m], {m, 1, Infinity}], 10] is better but takes several hours to compute, if it ever gives an solution!

The bottom line is the Mathematica code clearly shows (44) needs an overall minus in front of the summation.

Marvin Ray Burns
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Hmm, I don't know whether with "lucrative" it is actually meant "efficient"?

There is a possibility in Pari/GP to formulate the derivative of the eta() by a divergent sum-procedure sumalt(), applicable even to formal series. What I do in cases like this, where derivatives of eta at zero (or zeta at zero: Stieltjes-numbers) are involved is the following procedure 

default(seriesprecision,64)            \\ number of taylor-coefficients of a series

psaeta = sumalt(k=0,(-1)^k/(1+k)^x);   \\ gives the formal powerseries for aeta
pcaeta= polcoeffs(psaeta,64)~;         \\ extracts the taylor-coefficients
                                       \\ = eta^(k)(0) /k! 
   \\ now the functional definition for the c_j coefficients
mrbcj(j)=if(j==0,return(0));sum(d=1,j,binomial(j,d)*d^j/(-d)^d)          

mrb = sum(r=1,63,mrbcj(r)*pcaeta[1+r])   \\ this gives a rough estimate to about 4 digits

  \\ using Eulersummation I can improve the number of correct digits dramatically
ESum(1,64)*vectorv(64,r,mrbcj(r-1)*pcaeta[1+r])

The terms of the 60 to 64'th partial sums with the Euler-summation are

  ... ....
  -0.18785964246206711930
  -0.18785964246206712123
  -0.18785964246206712126
  -0.18785964246206712081

and the digits up to the last "7" seem to be correct. Perhaps there is something similar in Math'ca , at least they have manipulations for formal divergent series symbolically...

The Pari/GP-evaluation of the initial formula is very simple btw. and gives an accurate result very easy; I got 
    mrb =  sumalt(k=1,(-1)^k *(exp(log(k)/k)-1))
    \\ %2909 = 0.187859642462067120248517934054273230055903094900138786172005
Gottfried Helms
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