<s>[INST] Evaluate $\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} $, $p \in N$ 
Evaluate
  $$\lim_{n\to\infty}\frac{1}{n^{p+1}}\cdot \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} , p \in N$$ 

Now, I found this problem while doing some practice and I am curious on how to solve it . I  have no good ideas yet, so I will appreciate any hints !
 [/INST] A lower bound is given by
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\frac{{(p + i)!}}{{i!}}}  \ge \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {i^p } \\ = \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{n}\sum\limits_{i = 1}^n {\left( {\frac{i}{p}} \right)^p }  = \int_0^1 {x^p dx}  = \frac{1}{{p + 1}} .
$$
An upper bound:
\begin{align*}
& \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\frac{{(p + i)!}}{{i!}}}  = \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {(i + 1)(i + 2) \cdots (i + p)} 
\\ &
 \le \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\left( {i + \frac{{p + 1}}{2}} \right)^p }  \le \mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{n^{p + 1} }}\sum\limits_{i = 1}^n {\int_i^{i + 1} {\left( {x + \frac{{p + 1}}{2}} \right)^p dx} } 
\\ & =
\mathop {\lim }\limits_{n \to  + \infty } \frac{1}{{n^{p + 1} }}\int_1^{n + 1} {\left( {x + \frac{{p + 1}}{2}} \right)^p dx} \\ & = \frac{1}{{p + 1}}\mathop {\lim }\limits_{n \to  + \infty } \left( {1 + \frac{{p + 1}}{{2n}} + \frac{1}{n}} \right)^{p + 1}  - \frac{1}{{p + 1}}\mathop {\lim }\limits_{n \to  + \infty } \left( {\frac{{p + 1}}{{2n}} + \frac{1}{n}} \right)^{p + 1} 
\\ &
 = \frac{1}{{p + 1}}.
\end{align*}
I first used the inequality between the geometric and the arithmetic mean, then estimated each term by an integral taking into account the monotonicity of the power function. Thus, the limit in question is $\frac{1}{p+1}$.
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