A series with only rational terms for $\ln \ln 2$ We all know that 

$$
\ln 2 = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}.
$$

Do you know a series with only rational terms for

$$\ln \ln 2 = ?$$

Let's exclude base expansions with non explicit coefficients. 
Thanks!
 A: We have 

$$
\ln \ln 2  =   \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \! \left(\sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1}\right)\frac{1}{n!} \tag1
$$

where ${n \brack k}$ are the unsigned Stirling numbers of the first kind counting the numbers of permutations of
$n$ letters that have exactly $k$ cycles.
Proof.  Let $-1<t<1$. Recall the exponential generating function of the unsigned Stirling numbers of the first kind 
$$
\sum_{n=k}^{\infty} \displaystyle  \frac{{n \brack k}}{n!} t^{n} = \displaystyle \frac{\left(-\ln(1-t)\right)^{k}}{k!}, \quad k=1,2 \cdots. \tag2
$$
[Herbert S. Wilf, generatingfunctionology]   p. 82 (3.5.3)
On one hand we have
$$
\begin{align}
\displaystyle \sum_{n=0}^{\infty} (-1)^n\left( \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1}\right) \frac{t^{n-1}}{n!} & =  \frac{1}{t} + \frac{1}{t} 
 \sum_{k=1}^{\infty} \frac{1}{k+1} \sum_{n = k }^{\infty} \frac{{n \brack k}}{n!}(-t)^{n} \\
& =  \frac{1}{t} + \frac{1}{t} 
\sum_{k=1}^{\infty} \frac{\left(-\ln(1+t)\right)^{k}}{(k+1)!} \\
& =  \frac{1}{t} +
\displaystyle \frac{1}{-t\ln(1+t)}\left(e^{-\ln(1+t)} + \ln(1+t) - 1\right) \\
& = \frac{1}{(1+t) \ln(1+t)} ,
\end{align}
$$
that is
$$
\displaystyle \sum_{n=1}^{\infty} (-1)^n\left( \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1}\right) \frac{t^{n-1}}{n!} =   -\frac{1}{t}+ \frac{1}{(1+t) \ln(1+t)}. \tag3
$$
On the other hand we have
$$
\displaystyle \frac{d}{dt} \ln \left( \frac{\ln (1+t)}{t} \right)  =   -\frac{1}{t}+ \frac{1}{(1+t) \ln(1+t)}.  \tag4
$$
We deduce that

$$
\ln \left( \frac{\ln (1+t)}{t} \right)  =   \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \left( \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1}\right) \frac{t^{n}}{n!}, \quad 0<t<1. \tag5
$$

This gives, as $t$ tends to $1^-$, 
$$
\ln \ln 2  =   \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \! \left(\sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1}\right)\frac{1}{n!} .
$$
Remark. As $n$ tends to $\infty$, one may prove (here) that
$$
\frac{1}{n!} \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1} \sim \frac{1}{\ln n}. \tag6
$$
A: Such a rational series can certainly be written down, though I only consider this answer a partial result since I won't express it as a single sum. From the Taylor series $\ln(1+x)=-\sum_{j=1}^\infty (-x)^j/j$ we may write $\ln\ln(1+x)$ as
\begin{align}
 \ln\ln(1+x)
&= \ln\left(x-\sum_{j=2}^\infty\frac{(-x)^j}{j}\right)\\
&= \ln x + \ln\left(1+\sum_{j=2}^\infty\frac{(-x)^{j-1}}{j}\right) \\
&= \ln x-\sum_{k=1}^\infty\frac{1}{k}\left(-\sum_{j=2}^\infty\frac{(-x)^{j-1}}{j}\right)^k.
\end{align}
Note that the second term is certainly a rational series in powers of $x$, albeit not one which seems easy to express term-by-term; presumably one could expand using the binomial theorem and resum over $k$. The limit as $x\to 1$ then kills the first term and so yields
$$\ln\ln 2 = -\sum_{k=1}^{\infty}\frac{1}{k}\left(-\sum_{j=2}^\infty\frac{(-1)^{j-1}}{j}\right)^k.$$ Can anyone reduce this further? Note that this amounts to expressing the Taylor series of $\ln(\ln(1+x)/x)$ in a tractable way.

Update:
After some further work, I remembered that Faa di Bruno's formula expresses the Taylor series of $h(x)=f(g(x))$ in terms of Bell polynomials as $$h^{(n)}(x)=\sum_{k=1}^n f^{(k)}(g(x))\cdot B_{n,k}(g'(x),g''(x),\ldots,g^{(n-k+1)}(x).$$ We can use this to obtain the derivatives at $x=0$ in a rather mechanical way. For our purposes, we take $f(x)=\ln x$ and $g(x)=\ln(1+x)/x$. Then $f^{(k)}(x)=(-1)^{k-1} (k-1)!\,x^{-k}$ and $g^{(k)}(0)=(-1)^k k!/(k+1),$ so
\begin{align}
\frac{d^n}{dx^n}\left[\ln\left(\frac{\ln(1+x)}{x}\right)\right]_{x=0}
&=\sum_{k=1}^n f^{(k)}(1)\cdot B_{n,k}\left(-\frac{1}{2},\frac{2}{3},\ldots,(-1)^{n-k+1}\frac{(n-k+1)!}{n-k+2}\right) \\
&=\sum_{k=1}^n (k-1)!\,(-1)^{k-1} B_{n,k}\left(-\frac{1}{2},\frac{2}{3},\ldots,(-1)^{n-k+1}\frac{(n-k+1)!}{n-k+2}\right)
\end{align}
With this, we can express the Taylor series about $x=0$ for the function and take as above the limit as $x\to 1$:
\begin{align}
\ln\ln 2
&=\lim_{x\to 1^-}\ln\left[\frac{\ln(1+x)}{x}\right]
=\sum_{n=0}^\infty \frac{h^{(n)}(0)}{n!}\\
&=\sum_{n=0}^\infty \sum_{k=1}^n (-1)^{k-1} \frac{(k-1)!}{n!}\, B_{n,k}\left(-\frac{1}{2},\frac{2}{3},\ldots,(-1)^{n-k+1}\frac{(n-k+1)!}{n-k+2}\right).
\end{align}
This is still a pretty miserable looking expression, but this approach has gotten rid of the powers of $\ln(1+x)/x$ that were present above. Can anyone take this further?
