TITLE: What is the limit of $\lim_{n\to \infty} \frac{1}{n^{k+1}}\left(k!+\frac{(k+1)!}{1!}+\frac{(k+2)!}{2!}+\cdots+\frac{(k+n)!}{n!}\right)=?$
QUESTION [6 upvotes]: For $k\in\mathbb{N},$
$$\lim_{n\to \infty} \frac{1}{n^{k+1}}\left(k!+\frac{(k+1)!}{1!}+\frac{(k+2)!}{2!}+\cdots+\frac{(k+n)!}{n!}\right)=?$$  
So I am trying to find this limit. Try to apply limit comparison but don't know where to start. I observe that $(k+n)!\geq n!$, is this fact going to help me somehow? Help me out. Thanks.

REPLY [3 votes]: Move the $1/n^{k+1}$ inside the sum to see the expression equals
$$\tag1 S_n = \sum_{j=0}^{n} \frac{(j+k)!}{j!n^k}\frac{1}{n} = \sum_{j=0}^{n} \left(\frac{j}{n} + \frac{k}{n}\right)\left(\frac{j}{n} + \frac{k-1}{n}\right)\cdots \left(\frac{j}{n} + \frac{1}{n}\right)\frac{1}{n}.$$
Let $\epsilon>0.$ Then $k/n < \epsilon$ for large $n.$ For such $n,$ the right side of $(1)$ shows
$$\tag 2 \sum_{j=0}^{n} \left(\frac{j}{n}\right)^k\frac{1}{n} \le S_n \le \sum_{j=0}^{n} \left(\frac{j}{n}+\epsilon\right)^k\frac{1}{n}.$$
The left and right sums in $(2)$ are Riemann sums, and we see
$$\int_0^1 x^k\,dx \le \liminf S_n \le \limsup S_n \le \int_0^1 (x+\epsilon)^k\,dx.$$
As $\epsilon \to 0,$ the integral on the right converges to the integral on the left, which shows $\lim S_n = 1/(k+1).$