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 !
The limit is $\frac1{p+1}$. There is a nice closed form: $$ \sum_ \limits{i=1}^{n} \frac{(p+i)!}{i!} = \frac1{p+1} \underbrace{(n+1)\cdots (n+p+1)}_{\text{$p+1$ factors}} -p! $$ (even nicer if you absorb $p!$ into the LHS as the $i=0$ term.)
Proof 1. Telescoping! Write $$ \begin{align} \frac{(p+i+1)!}{i!}-\frac{(p+i)!}{(i-1)!}=\frac{(p+i)!}{i!}[(p+i+1)-i]=\frac{(p+i)!}{i!}(p+1). \end{align} $$ Sum from $i=1$ to $n$ to get $$ \frac{(p+n+1)!}{n!}-(p+1)! = (p+1)\sum_{i=1}^n\frac{(p+i)!}{i!}, $$ then divide through by $p+1$.
Proof 2: Use the Hockey Stick identity (H): $$ \frac1{p!}\sum_{i=0}^n\frac{(p+i)!}{i!}=\sum_{i=0}^n {p+i\choose p}=\sum_{t=p}^{p+n}{t\choose p}\stackrel{H}={p+n+1\choose p+1}=\frac1{(p+1)!}\frac{(p+n+1)!}{n!} $$ Multiply through by $p!$, and we're done.
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 $\frac{1}{p+1}$.
First of all, $ \left(\forall x\in\mathbb{R}_{+}\right),\ \mathrm{e}^{x}-1=x\int_{0}^{1}{\mathrm{e}^{xy}\,\mathrm{d}y}\leq x\, \mathrm{e}^{x} \cdot $
Let $ n,p $ be positive integers, we have the following : \begin{aligned} \left|\frac{1}{n^{p+1}}\sum_{i=1}^{n}{\frac{\left(p+i\right)!}{i!}}-\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}}\right|&=\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}\left(\prod_{j=1}^{p}{\left(1+\frac{j}{i}\right)}-1\right)}\\ &\leq\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}\left(\prod_{j=1}^{p}{\exp{\left(\frac{j}{i}\right)}}-1\right)}\\&\leq\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}\left(\exp{\left(\frac{p\left(p+1\right)}{2i}\right)}-1\right)}\\ &\leq\frac{1}{2n^{p+1}}\sum_{i=1}^{n}{i^{p-1}\exp{\left(\frac{p\left(p+1\right)}{2i}\right)}}\\ \left|\frac{1}{n^{p+1}}\sum_{i=1}^{n}{\frac{\left(p+i\right)!}{i!}}-\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}}\right|&\leq\frac{\mathrm{e}^{\frac{p\left(p+1\right)}{2}}}{2n}\times\frac{1}{n}\sum_{i=1}^{n}{\left(\frac{i}{n}\right)^{p-1}}\underset{n\to +\infty}{\longrightarrow}0\times\int_{0}^{1}{x^{p-1}\,\mathrm{d}x}=0 \end{aligned}
Thus $$ \lim_{n\to +\infty}{\frac{1}{n^{p+1}}\sum_{i=1}^{n}{\frac{\left(p+i\right)!}{i!}}}=\lim_{n\to +\infty}{\frac{1}{n^{p+1}}\sum_{i=1}^{n}{i^{p}}}=\int_{0}^{1}{x^{p}\,\mathrm{d}x}=\frac{1}{p+1} $$
Did you try by factoring the inside of summatory? Also, you can solve it for different values of p. (this should be a comment but i don't have enough reputation)
