What is the limit of $\lim_{n\to \infty} \frac{1}{n^{k+1}}\left(k!+\frac{(k+1)!}{1!}+\frac{(k+2)!}{2!}+\cdots+\frac{(k+n)!}{n!}\right)=?$ 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.
 A: 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 $\lim S_n = 1/(k+1).$
A: The sum in question is: 
$$ \frac{k!}{n^{k+1}} \left( \frac{k!}{k!} + \frac{(k+1)!}{k!1!} + \ldots + 
\frac {(k+n)!}{k!n!}\right).$$
Now the bracketed value is just $\binom k k + \binom {k+1} k + \ldots + 
\binom {k+n} k$. There's a well-known combinatorial identity that says this
is $\binom {k+n+1} {k+1}$ (quite easy to prove this by induction using $\binom a b + \binom a {b-1} = \binom {a+1} b$). Thus the above sum becomes:
$$ \frac{k!}{n^{k+1}} \binom {k+n+1}{k+1} = \frac{(n+1)(n+2)\ldots(n+k+1)}{(k+1)n^{k+1}} = \frac 1 {k+1}\left(1 + \frac 1 n\right)\ldots\left(1 + \frac {k+1}n\right).$$
As $n\to\infty$, this tends to $\frac 1 {k+1}$.
A: $$\lim_{n\to \infty} \frac{1}{n^{k+1}}\left(k!+\frac{(k+1)!}{1!}+\frac{(k+2)!}{2!}+\cdots+\frac{(k+n)!}{n!}\right)=$$
$$\lim_{n\to \infty} \frac{1}{n^{k+1}}\sum_{m=0}^{n}\frac{(k+m)!}{m!}=$$
$$\lim_{n\to \infty} \frac{1}{n^{k+1}}\cdot\frac{(n+1)(k+n+1)!}{(k+1)(n+1)!}=$$
$$\lim_{n\to \infty} \frac{1}{n^{k+1}}\cdot\frac{(k+n+1)!}{(k+1)n!}=$$
$$\lim_{n\to \infty} \frac{n^{-k-2}\Gamma(k+n+2)}{(k+1)\Gamma(n)}=$$
$$\frac{1}{k+1}\lim_{n\to \infty} \frac{n^{-k-2}\Gamma(k+n+2)}{\Gamma(n)}=\frac{1}{k+1}$$
