Evaluate limit $\lim_{n \to \infty } {1 \over n^{k + 1}}\left( {k! + {(k + 1)! \over 1!} + \cdots + {(k + n)! \over n!}} \right),k \in \mathbb{N}$ Evaluate the limit:
$$\lim_{n \to \infty } {1 \over n^{k + 1}}\left( {k! + {(k + 1)! \over 1!} + \cdots + {(k + n)! \over n!}} \right),k \in \mathbb{N}$$
It looks like a classic Cesaro-Stolz problem, but applying it didn't bring me any useful result. 
I've been told the following equality might be helpful: $(1 - q)(1 + q + \cdots + q^N) = (1 - q^{N + 1})$

Hint: It can be shown that
$$\sum_{m = 0}^n \frac{(k + m)!}{m!} = \frac{(n + 1)(k + n + 1)!}{(k + 1)(n + 1)!}.$$
You may find this formula useful.
If you then divide by $n^{k + 1}$ and take the limit as $n \to \infty$, you should get
$$\frac{1}{1 + k}.$$
Edit: Using Stirling's asymptotic formula $N! \sim N^N e^{-N}\sqrt{2\pi N}$, where $\sim$ denotes asymptotic equality, we have
$$\frac{(n + 1)(k + n + 1)!}{(n + 1)!} = \frac{(k + n + 1)!}{k^{k + 1}n!} \sim \frac{1}{n^{k + 1}} \frac{(k + n + 1)^{k + n + 1}e^{-(k + n + 1)}\sqrt{2\pi (k + n + 1)}}{n^n e^{-n}\sqrt{2\pi n}}.$$
If we simplify the right-hand side, we find that
\begin{align*}
\frac{1}{n^{k + n + 3/2}(k + n + 1)^{k + n + 3/2}e^{k + 1}} &= \left(1 + \frac{k + 1}{n}\right)^{-(k + n + 3/2)} e^{-(k + 1)}\\
&= \left(1 - \frac{-(k + 1)}{n}\right)^{-(k + n + 3/2)} e^{-(k + 1)}\\
&= \left[\left(1 - \frac{-(k + 1)}{n}\right)^{(k + n + 3/2)}\right]^{-1} e^{-(k + 1)}\\
&\to e^{k + 1}e^{-(k + 1)} \quad (\text{as } n \to \infty)\\
&= 1
\end{align*}
because
$$e^{-x} = \lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^n$$
and $k + n + 3/2 \sim n$ as $n \to \infty$ since $k \in \mathbb{N}$ is fixed.
