A series with only rational terms for $\ln \ln 2$

Question

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!

Answer 1

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.

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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?

Answer 2

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 $$ \frac{1}{n!} \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1} \sim \frac{1}{\ln n}. \tag6 $$