limit of a sequence. might be related to Cesaro theorem this it the limit to evaluate:
$$\mathop {\lim }\limits_{n \to \infty } {1 \over {{n^{k + 1}}}}(k! + {{(k + 1)!} \over {1!}} + ... + {{(k + n)!} \over {n!}})$$
I've given an hint which is:
$$(1 - q)(1 + q + ... + {q^N}) = (1 - {q^{N + 1}})$$
I tried to use Cesaro theorem but got to sort of a dead end..
I also noticed a similarity to the Binomial coefficient, is that going to help me?
Any help will be appreciated
 A: We can write the sum as a rising sum of binomial coefficients in the following manner:
$$\sum_{r=0}^n \frac{(k+r)!}{r!} = k! \sum_{r=0}^n \binom{k+r}{k} = k! \binom{n+k+1}{n} = \frac{(n+k+1)!}{n!(k+1)}$$
So the desired limit is 
$$\frac1{k+1}\lim_{n \to \infty}\frac{(n+k+1)!}{n^{k+1}\, n!} = \frac1{k+1}\lim_{n \to \infty}\left(\frac{n+1}{n}\cdot\frac{n+2}{n}\cdot\frac{n+3}{n}\dots \frac{n+k+1}{n} \right) = \frac1{k+1}$$
A: So, we get $$\frac{\left((k)(k-1)...(1)+...+(n+k)(n+k-1)...(n+1)\right)}{n^{k+1}}.$$ 
The sum in brackets is a linear combination (with constant coefficients, constant in $n$) of the sums of the $r$ powers of the first $n$ numbers, for $r=0,...,k$.
$$\frac{\left(\sum_{r=1}^{n}r^k+C_1(k)\sum_{r=1}^{n}r^{k-1}+...+C_{k}(k)\sum_{r=1}^{n}1\right)}{n^{k+1}}.$$
All of them, except the $k$-th powers are polynomials of degree less than $k+1$. So their limits, when divided by $n^{k+1}$, go to zero. 
So, the limit is the same as $$\lim_{n\rightarrow\infty}\frac{1^k+2^k+...+n^k}{n^{k+1}}$$.
The sum of the $k$-th powers is a polynomial of degree $k+1$ and with leading coefficient $\frac{1}{k+1}$. So, that is your limit.
A: Stirling's formula tells us that
$$
\frac{(k+m)!}{m!} \sim m^k
$$
as $m \to \infty$, so, by Stolz-Cesàro,
$$
\sum_{m=1}^{n} \frac{(k+m)!}{m!} \sim \sum_{m=1}^{n} m^k
$$
as $n \to \infty$.  Now
$$
\int_0^n x^k\,dx \leq \sum_{m=1}^{n} m^k \leq n^k + \int_0^n x^k\,dx
$$
and consequently
$$
\sum_{m=1}^{n} m^k \sim \frac{n^{k+1}}{k+1}
$$
as $n \to \infty$.  Thus
$$
\lim_{n \to \infty} \frac{1}{n^{k+1}} \sum_{m=1}^{n} \frac{(k+m)!}{m!} = \lim_{n \to \infty} \frac{1}{n^{k+1}} \cdot \frac{n^{k+1}}{k+1} = \frac{1}{k+1}.
$$
