TITLE: An asymptotic term for a finite sum involving Stirling numbers
QUESTION [9 upvotes]: The question is a by-product at the end of this post.
The following asymptotic term will ensure the convergence of some series.

$$
\frac{1}{n!} \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1} = \mathcal{O} \left(\frac{1}{\ln n}\right), \quad \text{as} \quad n \rightarrow \infty.
$$

Here $\displaystyle {n \brack k}$ are the unsigned Stirling numbers of the first kind.
Can you find a proof for it?
Update: A related interesting paper may be found here.

REPLY [3 votes]: I computed this asymptotic expansion for this answer. To relate the two, we will use
$$
\newcommand{\stirone}[2]{\left[{#1}\atop{#2}\right]}
\sum_{k=0}^n\stirone{n}{k}x^k=\frac{\Gamma(x+n)}{\Gamma(x)}\tag{1}
$$
Integrating $(1)$ on $[0,1]$ and dividing by $n!=\Gamma(n+1)$ gives
$$
\frac1{n!}\sum_{k=0}^n\frac{\stirone{n}{k}}{k+1}
=\int_0^1\frac{\Gamma(n+x)}{\Gamma(n+1)\Gamma(x)}\mathrm{d}x\tag{2}
$$
For $x\in[0,1]$, Gautschi's Inequality says
$$
(n+1)^{x-1}\le\frac{\Gamma(n+x)}{\Gamma(n+1)}\le n^{x-1}\tag{3}
$$
Dividing $(3)$ by $\Gamma(x)$ and integrating on $[0,1]$ gives
$$
\int_0^1\frac{(n+1)^{x-1}}{\Gamma(x)}\mathrm{d}x
\le\int_0^1\frac{\Gamma(n+x)}{\Gamma(n+1)\Gamma(x)}\mathrm{d}x
\le\int_0^1\frac{n^{x-1}}{\Gamma(x)}\mathrm{d}x\tag{4}
$$
Substituting $x\mapsto1-x$, we get the asymptotic expansion
$$
\begin{align}
\int_0^1\frac{n^{x-1}}{\Gamma(x)}\mathrm{d}x
&=\int_0^1\frac{n^{-x}}{\Gamma(1-x)}\mathrm{d}x\\
&=\int_0^1\left(1-\gamma x-\left(\tfrac{\pi^2}{12}-\tfrac{\gamma^2}2\right)x^2+O\left(x^3\right)\right)e^{-x\log(n)}\,\mathrm{d}x\\
&=\frac1{\log(n)}-\frac\gamma{\log(n)^2}-\frac{\frac{\pi^2}6-\gamma^2}{\log(n)^3}+O\left(\frac1{\log(n)^4}\right)\tag{5}
\end{align}
$$
Since the difference between $(5)$ for $n$ and $n+1$ is $O\left(\frac1n\right)$, $(4)$ says we have
$$
\int_0^1\frac{\Gamma(n+x)}{\Gamma(n+1)\Gamma(x)}\mathrm{d}x=\frac1{\log(n)}-\frac\gamma{\log(n)^2}-\frac{\frac{\pi^2}6-\gamma^2}{\log(n)^3}+O\left(\frac1{\log(n)^4}\right)\tag{6}
$$