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\markboth{}{{\sc  Answer to Question 1 in Zeilbergers' `Fractional Counting of Partitions' }}

\bigskip
This is a short solution of Question 1 posed in the article
\bigskip
{\it Fractional Counting of Partitions} by Doron and Noam Zeilberger
\bigskip

See {\tt http://sites.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/fcp.html}  and
\bigskip

{\tt https://arxiv.org/abs/1810.12701} \quad .
\bigskip
The first solution was given by Will Sawin, and is included as an appendix in the current version of their paper.
\bigskip
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\[
 \textit{If}\quad C:=\lim \frac{b(n)}n \quad\textit{exists then}\quad C=e^{-\gamma}.
\]


Given $\epsilon>0$, $|b(n)-Cn|<\epsilon n$ for $n$ large enough. Hence there exists $\delta<1$ such that if $B(z)$ is the generating function of $b(n)$ and $\phi(z)=z/(1-z)^2$ (the generating function of $n$), we have
$|B(x)-C\phi(x)|<\epsilon\phi(x)$ for $1-\delta<x<1$. Multiplying by $(1-x)^2$, $C=\lim_{x\to 1^-} (1-x)^2B(x)$. 
Taking logarithms
\[
 \log C
 =
 \lim_{x\to 1^-}
 \Big(
 \log(1-x)
 -
 \sum_{k=2}^\infty
 \log\big(
 1-\frac{x^k}{k}
 \big)
 \Big)
 =
 -\lim_{x\to 1^-}
 \Big(
 x
 +
 \sum_{k=2}^\infty
 \Big(
 \log\big(
 1-\frac{x^k}{k}
 \big)
 +\frac{x^k}{k}
 \Big)
 \Big).
\]
Hence
\[
 \log C
 =
 -1
 -
 \sum_{k=2}^\infty
 \Big(
 \frac 1k
 -
 \log\frac{k}{k-1}
 \Big)
 =
 -
 \lim_{N\to \infty}
 \sum_{k=1}^N
 \Big(
 \frac 1k
 -\log N
 \Big)
 =-\gamma. 
\]

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\

Fernando Chamizo.  Department of Mathematics, Universidad Aut\'onoma de Madrid and
ICMAT, 28049 Madrid, Spain.

E-mail address: \url{fernando.chamizo@uam.es}



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