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This article has been edited for a bounty.

$C$ MRB, the MRB constant, is defined at http://mathworld.wolfram.com/MRBConstant.html .

There is an excellent 56 page paper whose author has passed away. You can find it in Google Scholar "MRB constant," Better yet, use the following link http://web.archive.org/web/20130430193005/http://www.perfscipress.com/papers/UniversalTOC25.pdf. You find a cached copy there.

Just before the author, Richerd Crandall, died I wrote him about a possible small error. What I'm worried about is formula 44 on page 29 and below. When I naively worked formula 44 it needed a negative sign in front of it. Crandall did write me back admitting to a typo, but he died before he had a chance to correct it. Is there anyone out there competent enough to check, correct and prove the corrected formulas for me? Thank you. I will use the proofs often and try to get the formulas published more.

part1

part1 Here is how I worked formula 44 and got -B:

(*define the Dirichlet 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]

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  • $\begingroup$ Would it be possible for you to type in the formula from the paper, and indicate the suspected typo? I don't care to download a 56(-page?) paper just to check one formula. $\endgroup$ – Gerry Myerson Nov 2 '13 at 23:36
  • $\begingroup$ @Gerry Myerson , I added the formula $\endgroup$ – Marvin Ray Burns Nov 3 '13 at 14:23
  • $\begingroup$ Thanks. Not enough info there to tell --- in particular, you'd have to know what $\eta$ stands for. Sorry --- I hope someone else will download the paper, and give it a try. $\endgroup$ – Gerry Myerson Nov 3 '13 at 22:10
  • $\begingroup$ That doesn't tell me what $\eta$ stands for. $\endgroup$ – Gerry Myerson Nov 4 '13 at 2:39
  • 3
    $\begingroup$ The context indicates it is the Dirichlet eta function. $\endgroup$ – anon Nov 5 '13 at 16:03
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The first equality can be reproduced by formally stating the double sum: write down the expanded exponantial series for each term in one row, sum then column-wise to get the derivatives of the $\eta()$ and sum then the $\eta()$-expressions. Here the $\eta(s)$ is the alternating $\zeta(s)$ and $\eta^{(m)}(s)$ the $m$'th derivative.

$$ \begin{array} {rclll} -\exp( {\log(1) \over 1})+1 & = & -{\log(1) \over 1} &-{\log(1)^2 \over 1^2 2!} &-{\log(1)^3 \over 1^3 3!} & - \cdots \\ +\exp( {\log(2) \over 2})-1 & = &+{\log(2) \over 2} &+{\log(2)^2 \over 2^2 2!} &+{\log(2)^3 \over 2^3 3!} & + \cdots \\ -\exp( {\log(3) \over 3})+1 & = &-{\log(3) \over 3} &-{\log(3)^2 \over 3^2 2!} &-{\log(3)^3 \over 3^3 3!} & - \cdots \\ \vdots \qquad & \vdots & \quad\vdots & \quad\vdots& \quad\vdots & \ddots \\ \hline \\ B \qquad & = & {\eta^{(1)}(1) \over 1!} &- {\eta^{(2)}(2) \over 2!} &+ {\eta^{(3)}(3) \over 3!} & - \cdots \end{array}$$

I've not yet the expansion into the formula with the derivatives of $\eta()$ at zero; perhaps I can supply that tomorrow.


Numerical check

In Pari/GP I get for the original sum $$ B = \sum_{k=1}^\infty (-1)^k( \exp( \log(k)/k)-1)$$

                           B=sumalt(k=1,(-1)^k * (exp( log(k)/k) -1)      

well converging the value

       B=0.187859642462067120248517934054273230055903094900138786172005...     

as well for

                         B = -sum(k=1,6,(-1)^k*aetad(k,k)/k!)          

(but to fewer digits).
My user-defined function aetad(s,d) gives the d'th derivative of $\eta()$ at $s$ and I implemented that aeta() using the $\eta()$ / $\zeta()$ conversion formula

              aeta(s) = if(s==1,return(log(2));return( (1-2/2^s)*zeta(s)  )
              {aetad(s=1,d=0)=local(h1=1e-12,z);          
                      z= sum(k=0,d,  (-1)^k*binomial(d,k)*aeta(s +(d/2-k)*h1));
                         return( z/h1^d); }


I have also another implementation of the $\eta()$ and its derivatives which allows to access higher derivatives (in the above version the 10'th or even 20'th derivative were impractical). This version is also independent of the software-included zeta-function but uses Pari/GP's sumalt() procedure which allows to evaluate alternating (moderately) divergent series. My approach computes the taylor-expansion for the $\eta()$ function by default centered around zero but can easily be generalized to any center:

fmt(200,12)  \\ user defined; set internal precision to 200 digits, display to 12 digits
default(seriesexpansion,24)          \\ set expansion-limit for power series
m_coeffs = vectorv(24);              \\ vector for 24 aeta-taylor-series
{for(m=0,23,
    tay_aeta = sumalt(k=0,(-1)^k/(1+k)^(x+m));  \\ taylorseries for aeta around m
    m_coeffs[1+m]= polcoeffs(tay_aeta,24);      \\ store leading 24 taylorcoefficients
   );
  m_coeffs = Mat(m_coeffs);}        \\ make a matrix from the list of taylor series

After that, the desired coefficients (the derivatives of the $\eta(m)$ ) nicely divided by the factorials, occur on the diagonal of the coefficients-matrix, and we only need sum them with alternating sign:

   B = sum (k=1,23, -(-1)^k*m_coeffs[1+k,1+k])
   \\ 0.1878596424620671202485179340542732... the leading correct digits
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