Is there a more rigorous way to show these two sums are exactly equal? I would like to have a more rigorous proof of the hypothesis: The Crandall eta derivative series is equal to the following more elementary one.
The following two series give the same sum.
They are 
$$-\sum _{n=1}^{\infty } \frac{(-1)^n  \eta ^{(n)}(n)}{n!}$$ 
and
$$\sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{1}{n}}-1\right).$$
@Gottfried Helms 
made the following proof that is correct as far as the arithmetic and numeric evaluation of the eta derivatives from the logs. However, even as a novice, I think it lacks any formal proving power to show, logarithms do add up to exactly the eta derivatives! I think he was just pointing out to me that there is some trivialness in equating the two sums. 

[sic]

By use a series for
$n^{1/n} = \exp(\log n/n)$ and laying the two series: $$\sum _{n=1}^{\infty } (-1)^n \left(n^{\frac{1}{n}}-1\right)$$ and $$-\sum _{n=1}^{\infty } \frac{(-1)^n  \eta ^{(n)}(n)}{n!}$$
 out in two dimensions:
$$ \begin{array} {rclll} 
 -\exp( {\log(1) \over 1})+1 & = & -{\log(1) \over 1} 
                                   &-{\log(1)^2 \over 1^2 2!}
                                   &-{\log(1)^3 \over 1^3 3!} & - \cdots \\
 +\exp( {\log(2) \over 2})-1 & = &+{\log(2) \over 2} 
                                   &+{\log(2)^2 \over 2^2 2!}
                                   &+{\log(2)^3 \over 2^3 3!} & + \cdots \\
 -\exp( {\log(3) \over 3})+1 & = &-{\log(3) \over 3} 
                                   &-{\log(3)^2 \over 3^2 2!}
                                   &-{\log(3)^3 \over 3^3 3!} & - \cdots \\
    \vdots \qquad                  & \vdots & \quad\vdots & \quad\vdots& \quad\vdots
         & \ddots \\
\hline \\
    B \qquad & = &   {\eta^{(1)}(1) \over 1!} 
                  &- {\eta^{(2)}(2) \over 2!} 
                  &+ {\eta^{(3)}(3) \over 3!} & - \cdots
\end{array}$$
Can you improve his partial proof? 
 A: Even though one has cause to be a little bit wary around formal rearrangements of conditionally convergent sums (see the Riemann series theorem), it's not very difficult to validate the formal manipulation of Helms. The idea is to cordon off a big chunk of the infinite double summation (all the terms from the second column on) that we know is absolutely convergent, which we are then free to rearrange with impunity. (Most relevantly for our purposes here, see pages 80-85 of this document, culminating with the Fubini theorem which is essentially the manipulation Helms is using.) 
So, by definition the MRB constant $B$ is the conditionally convergent sum $\sum_{n \geq 1} (-1)^n (n^{1/n} - 1)$. Put $a_n = (-1)^n (n^{1/n} - 1)$, so $B = \sum_{n \geq 1} a_n$. Looking at the first column, put $b_n = (-1)^n \frac{\log(n)}{n}$, so $\eta^{(1)}(1) = \sum_{n \geq 1} b_n$ as a conditionally convergent series. 
We have 
$$B - \eta^{(1)} = \sum_{n \geq 1} a_n - b_n = \sum_{n \geq 1} \sum_{m \geq 2} (-1)^n \frac{(\log n)^m}{n^m m!}$$ 
(The first equation is an elementary limit statement that says if $\sum_{n \geq 1} a_n$ converges and $\sum_{n \geq 1} b_n$ converges, then also $\sum_{n \geq 1} a_n - b_n$ converges and $\sum_{n \geq 1} a_n - \sum_{n \geq 1} b_n = \sum_{n \geq 1} a_n - b_n$. It doesn't at all matter whether the convergence of either series is conditional or absolute.) 
So now we check the absolute convergence of the right-hand side, i.e., that $\sum_{n \geq 1} \sum_{m \geq 2} \frac{(\log n)^m}{n^m m!}$ converges. (Remember what this means in the case of infinite sums of positive terms: it means that there is a number $K$ such that every finite partial sum $S$ is bounded above by $K$; the least such upper bound will be the number that the infinite sum converges to.) So take any such finite partial sum $S$, and rearrange its terms so that the terms in the $m = 2$ column come first, then the terms in the $m = 3$ column, and so on. An upper bound for the terms of $S$ in the $m = 2$ column is $\frac{\zeta^{(2)}(2)}{2!}$. Put that one aside. 
For the $m = 3$ column, an upper bound is $\sum_{n \geq 2} \frac{(\log n)^3}{n^3 3!}$ (we drop the $n=1$ term which is $0$). By calculus we have $\log n \leq n^{1/2}$ for all $n \geq 2$, so this has upper bound $\frac1{3!} \sum_{n \geq 2} \frac1{n^{3/2}} \leq \frac1{3!} \int_1^\infty \frac{dx}{x^{3/2}}$ by an integral test, which yields $\frac{2}{3!}$ as an upper bound. Applying the same reasoning for the $m$ column from $m = 4$ on, an upper bound for that column would be $\frac1{m!} \int_1^\infty \frac{dx}{x^{m/2}} = \frac{2}{m!(m-2)}$. Adding all those upper bounds together, an upper bound for the entire doubly infinite sum would be 
$$\frac{\zeta^{(2)}(2)}{2!} + \sum_{m \geq 3} \frac{2}{m!(m-2)}$$ 
which certainly converges. So we have absolute convergence of the doubly infinite sum. 
Thus we are in a position to apply the Fubini theorem, which justifies the rearrangement expressed in the first of the following equations 
$$\sum_{n \geq 1} \sum_{m \geq 2} (-1)^n \frac{(\log n)^m}{n^m m!} = \sum_{m \geq 2} \sum_{n \geq 1} (-1)^n \frac{(\log n)^m}{n^m m!} = \sum_{m \geq 2} (-1)^{m+1} \frac{\eta^{(m)}(m)}{m!}$$ 
giving us what we wanted. 
