An asymptotic term for a finite sum involving Stirling numbers 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 foundhere.**

We have:

$$\sum_{k=0}^n {n\brack k}x^k=x(x+1)\cdots(x+n-1)$$ (See Comtet, Advanced combinatorics) and $$u_n=\frac{1}{n!} \sum_{k = 1 }^{n } \frac{{n \brack k}}{k+1}=\int_0^1\frac{t(t+1)\cdots (t+n-1)}{n!} dt$$

Now we have $\exp(-x)\geq 1-x$ for $x\geq 0$. Hence for $t\in [0,1]$ we have $$\frac{t(t+1)\cdots (t+n-1)}{n!}=t\prod_{k=1}^{n-1} (1-\frac{1-t}{k+1})\leq t\exp(-v_n (1-t))$$ where $1+v_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$.

Hence $$u_n\leq \int_0^1 t\exp(-v_n (1-t))=\frac{1}{v_n}-\frac{1-\exp(-v_n)}{v_n^2}\leq \frac{1}{v_n}$$

As $v_n\sim \log n$, we are done.