TITLE: A series with only rational terms for $\ln \ln 2$
QUESTION [13 upvotes]: 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!

REPLY [6 votes]: 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
$$
\frac{1}{n!} \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1} \sim \frac{1}{\ln n}. \tag6
$$