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: $(1 - q)(1 + q + \cdots + q^N) = (1 - q^{N + 1})$
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$\ds{\lim_{n \to \infty }{1 \over n^{k + 1}}\bracks{%
     k! + {\pars{k + 1}! \over 1!} + \cdots + {\pars{k + n}! \over n!}}\,,\quad
     k \in \mathbb{N}:\ {\large ?}}$

\begin{align}
&\lim_{n \to \infty }{1 \over n^{k + 1}}\bracks{%
k! + {\pars{k + 1}! \over 1!} + \cdots + {\pars{k + n}! \over n!}}
=\lim_{n \to \infty}{1 \over n^{k + 1}}
\sum_{\ell = 0}^{n}{\pars{k + \ell}! \over \ell!}
\\[3mm]&=k!\,\lim_{n \to \infty}{1 \over n^{k + 1}}
\sum_{\ell = 0}^{n}{k + \ell \choose k}
=k!\,\lim_{n \to \infty}{1 \over n^{k + 1}}
\sum_{\ell = 0}^{n}\int_{\verts{z} = 1}{\pars{1 + z}^{k + \ell} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=k!\,\lim_{n \to \infty}{1 \over n^{k + 1}}
\int_{\verts{z} = 1}{\pars{1 + z}^{k} \over z^{k + 1}}
\sum_{\ell = 0}^{n}\pars{1 + z}^{\ell}\,{\dd z \over 2\pi\ic}
\\[3mm]&=k!\,\lim_{n \to \infty}{1 \over n^{k + 1}}
\int_{\verts{z} = 1}{\pars{1 + z}^{k} \over z^{k + 1}}\,
{\pars{1 + z}^{n + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}
\\[3mm]&=k!\,\lim_{n \to \infty}{1 \over n^{k + 1}}
\bracks{%
\overbrace{%
\int_{\verts{z} = 1}{\pars{1 + z}^{k + n + 1} \over z^{k + 2}}
\,{\dd z \over 2\pi\ic}}^{\ds{=\ {k + n + 1 \choose k + 1}}}\
-\
\overbrace{%
\int_{\verts{z} = 1}{\pars{1 + z}^{k} \over z^{k + 2}}
\,{\dd z \over 2\pi\ic}}^{\ds{=\ 0}}}
\\[3mm]&=k!\,
\lim_{n \to \infty}{1 \over n^{k + 1}}\,{\pars{k + n + 1}! \over \pars{k + 1}!\,n!}
\\[3mm]&={k! \over \pars{k + 1}!}\,
\lim_{n \to \infty}{1 \over n^{k + 1}}\,
{\root{2\pi}\pars{k + n + 1}^{k + n + 3/2}\expo{-n - k - 1} \over
\root{2\pi}n^{n + 1/2}\expo{-n}}
\\[3mm]&={1 \over k + 1}\,
\overbrace{\lim_{n \to \infty}\pars{1 + {k + 1 \over n}}^{k + n + 3/2}\expo{-k - 1}}
^{\ds{=\ 1}}
\end{align}

$$\color{#00f}{\large%
\lim_{n \to \infty }{1 \over n^{k + 1}}\bracks{%
k! + {\pars{k + 1}! \over 1!} + \cdots + {\pars{k + n}! \over n!}}
={1 \over k + 1}}
$$
A: 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.
A: Factor out the k!. Use the identity: 1 + C(k,k) + C(k+1,k) +...+ C(k+n,k) = C(k+n+1,k+1). Thus the expression equals: k!(n+k+1)!/((k+1)!n!n^(k+1)) = (1+1/n)(1+2/n)...(1+(k+1)/n)/(k+1) ---> 1/(k+1) as n --> infinity.
