What is $\sum_{n} |S_{1} (n,k)| x^{n} $?

Question

I'm trying to find out what the generating sum in the question title amounts to. So far, I've found the following related results (see p. 5 of this paper):

$$ \sum_{l=1}^{n} |S_{1}(n,l)|z^{l} = (z)_{n} = \prod_{k=0}^{n-1} (z+k) , $$ which amounts to $n!$ when $z=1$, and $$ \sum_{n=l}^{\infty} \frac{|S_{1}(n,l)|}{n!}z^{n} = (-1)^{l} \frac{\ln^{l}(1-z)}{l!} .$$ Here, $S_{1}(n,l)$ are the Stirling numbers of the first kind, without sign. In Generatingfunctionology by Wilf, I've also found that $$\sum_{k} |S_{1}(n,k)|y^{k} = n! \binom{y+n-1}{n} .$$ Instead of summing the above expression over $n$, I wish to find the sum over $n$. On p. 73 of Wilf's book, we see that $$ |S_{1} (n,k)| = \Big{[}\frac{x^{n}}{n!} \Big{]} \frac{1}{k!} \Big{(} \log \frac{1}{1-x} \Big{)}^{k} .$$

Therefore, I tried the following: \begin{align*} \sum_{n} |S_{1} (n,k)| x^{n} &= \sum_{n} x^{n} \Big{[}\frac{x^{n}}{n!} \Big{]} \frac{1}{k!} \Big{(} \log \frac{1}{1-x} \Big{)}^{k} \\ &= \frac{1}{k!} \Big{(} \log \frac{1}{1-x} \Big{)}^{k} \Big{[}\frac{x^{n}}{n!} \Big{]} \sum_{n} x^{n} \\ &= \frac{1}{k!} \Big{(} \log \frac{1}{1-x} \Big{)}^{k} \Big{[}\frac{x^{n}}{n!} \Big{]} \frac{1}{1-x} . \end{align*}

Questions:

1. Is what I've done so far correct? If not, why not? And how can the derivation be improved?
2. If it is correct, then what is $\Big{[}\frac{x^{n}}{n!} \Big{]} \frac{1}{1-x} $ ?
3. Is there a reference somewhere of the correct result?

Answer 1

There can be no closed form for $\sum_n |S_1(n,k)|x^n$, since this is infinite for all $x> 0$. The problem is that $S_1(n,k)$ grows too quickly; it is on the order of $n!$ as $n\to\infty$. This is fixed by dividing by $n!$ in the $\sum_n |S_1(n,k)|x^n/n!$ result you mentioned, but without the $n!$, there can be no solution.

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There is an error where you are using the same variable $x$ to mean two different things. To be correct, you should write $$ \sum_n |S_1(n,k)|x^n=\sum_n x^n\left[\color{blue}{t}^n \over n!\right]\frac1{k!}\left(\log\frac1{1-\color{blue}t}\right)^k $$ Then, using the fact that $[t^n/n!]f(t)=n![t^n] f(t)$, and the known Maclaurin series for $\log\frac1{1-t}$, this becomes $$ \sum_n |S_1(n,k)|x^n=\sum_n x^n\frac{n!}{k!}[t^n]\left(\log\frac1{1-\color{black}t}\right)^k $$ However, there is no nice way to rewrite $[t^n]\left(\log\frac1{1-t}\right)^k$ without using $S_1(n,k)$, so this is a dead end.