Jtdylktuy

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.

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]

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.

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

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

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); }

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