The step I'm having trouble with is
$\sum _{n=2}^{\infty } (-1)^n \left(n^{1/n}+\eta '(n)-\frac{\log (n)}{n-1}-1\right)+\eta '(1) =-\left(\sum _{m=2}^{\infty } \frac{(-1)^m \eta ^m(m)}{m!}+\eta '(1)\right)$
Expounding upon the MRB constant, I found the following equation, where I go in steps from a single conditionally to a single absolutely convergent sequence. I can compute and analyze each line using Mathematica, but I would like to present something more elegant than that at my discussion in https://community.wolfram.com/groups/-/m/t/366628.
(In part, I added and subtracted the following conditionally convergent series, $$\sum _{n=1}^{\infty } \eta'(n) =\sum _{n=2}^{\infty } (-1)^n \left(\frac{\log (n)}{n-1}\right) $$ but I thought there would be a rigorous way to show their equality is correct. I also combined that with the comment I got in $ C_{MRB}=\sum _{x=1}^{\infty } (-1)^x\left(e^{\frac{\log x}{x}}-1\right)$ Is its absolutely convergent arrangement prime number theorem related? ,
You know the terms of your series tend to zero. But how fast? Setting $u=\frac{\log x}{x}$, you have $(e^u-1)\sim u$ tends to $0$ "linearly." And $(e^u-1-u)\sim\frac{1}{2}u^2$ tends to $0$ "quadratically." Linear doesn't give us absolute convergence the way quadratic does. How did they know to subtract $u$ from $e^u-1$ to get quadratic decay? Well, because $e^u=1+u+\frac{1}{2}u^2+\cdots$ is the Taylor series for $e^u$ around $u=0$. (The series $\sum_{x=1}^\infty(-1)^x\frac{\log x}{x}=\gamma(\log2)-\frac{1}{2}(\log 2)^2$ they carved out of the original is itself conditionally convergent, BTW),
where I repeated the carving out indefinitely to get the term, $\frac{\log (n)}{n-1}.$ used above.)
$$\sum _{n=1}^{\infty } (-1)^n\left(n^{1/n}-1\right)$$ $$=\sum _{n=2}^{\infty } (-1)^n \left(n^{1/n}-\frac{\log (n)}{n-1}-1\right)+\sum _{n=1}^{\infty } \eta'(n) $$ $$=\sum _{n=2}^{\infty } (-1)^n \left(n^{1/n}+\eta '(n)-\frac{\log (n)}{n-1}-1\right)+\eta '(1)$$ $$=-\left(\sum _{m=2}^{\infty } \frac{(-1)^m \eta ^m(m)}{m!}+\eta '(1)\right)$$ $$=-\sum _{m=1}^{\infty } \frac{(-1)^m \eta ^m(m)}{m!}$$
$ \eta(m)$ is the Dirichlet eta function
Here is most of the work I did "where I repeated the carving out indefinitely:" 
Gottfried Helms gave this beautiful proof to get the last sum from the first, but I would like to see it proved step by step.
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.
$$ \begin{array} {rclll} -\exp( {\log(1) \over 1})+1 & = & -{\log(1) \over 1} &-{\log(1)^2 \over 1^2 2!} &-{\log(1)^3 \over 1^3 3!} & - \cdots \\ +\exp( {\log(2) \over 2})-1 & = &+{\log(2) \over 2} &+{\log(2)^2 \over 2^2 2!} &+{\log(2)^3 \over 2^3 3!} & + \cdots \\ -\exp( {\log(3) \over 3})+1 & = &-{\log(3) \over 3} &-{\log(3)^2 \over 3^2 2!} &-{\log(3)^3 \over 3^3 3!} & - \cdots \\ \vdots \qquad & \vdots & \quad\vdots & \quad\vdots& \quad\vdots & \ddots \\ \hline \\ B \qquad & = & {\eta^{(1)}(1) \over 1!} &- {\eta^{(2)}(2) \over 2!} &+ {\eta^{(3)}(3) \over 3!} & - \cdots \end{array}$$
Here is the best proof I have. There could be some errors in it.
$$\sum _{n=1}^{\infty } (-1)^n\left(n^{1/n}-1\right) \\ =\sum _{n=2}^{\infty } (-1)^n \left(n^{1/n}-\frac{\log (n)}{n-1}-1\right)+\sum _{n=1}^{\infty } \eta'(n) \\ =\sum _{n=2}^{\infty } (-1)^n \left(n^{1/n}+\eta '(n)-\frac{\log (n)}{n-1}-1\right)+\eta '(1)$$
\begin{array} {rclll} \eta'(1) & = & -{\log(1) \over 1} &+{\log(1)^2 \over 2} &-{\log(1)^3 \over 3} & + \cdots \\ +2^{1/2}+\eta'(2)-{\log(n)\over(n-1)}-1 & = & a2+{\log(2) \over 2} &-{\log(2)^2 \over 2^2 2!} &+{\log(2)^3 \over 2^3 3!} & - \cdots \\ -3^{1/3}-\eta'(3)+{\log(n)\over(n-1)}+1 & = &a3-{\log(3) \over 3} &+{\log(3)^2 \over 3^2 2!} &-{\log(3)^3 \over 3^3 3!} & + \cdots \\ +4^{1/4}+\eta'(4)-{\log(n)\over(n-1)}-1 & = &a4+{\log(4) \over 4} &{\log(4)^2 \over 4^2 2!} &+{\log(4)^3 \over 4^3 3!} & - \cdots \\ \vdots \qquad & \vdots & \quad\vdots & \quad\vdots& \quad\vdots & \ddots,\text{ } {a2+a3+a4+...=0}\\ \hline \\ CMRB \qquad & = & +{\eta^{(1)}(1) \over 1!} &- {\eta^{(2)}(2) \over 2!} &+ {\eta^{(3)}(3) \over 3!} & - \cdots \end{array} \ $$=-\sum _{m=1}^{\infty } \frac{(-1)^m \eta ^m(m)}{m!}$$