TITLE: Evaluate $(1+p)\cdots(1+\frac pn)$
QUESTION [5 upvotes]: Prove for $\forall p\in\mathbb{R}$, 
  $$\lim_{n\rightarrow\infty}\dfrac{\prod_{k=1}^n(1+\frac{p}{k})}{n^p}=L_p$$
  where $L_p\in\mathbb{R}$.

Moreover, I tried to calculate $L_p$, and it can be calculated if $p$ is an integer. What if $p$ is not an integer?

REPLY [2 votes]: Since the result is
$\dfrac1{\Gamma(p+1)}$,
I'll try to get
$\dfrac1{p!}$
for integer $p$
in an relatively elementary way.
$\begin{array}\\
L(p, n)
&=\dfrac{\prod_{k=1}^n(1+\frac{p}{k})}{n^p}\\
&=\dfrac{\prod_{k=1}^n(\frac{k+p}{k})}{n^p}\\
&=\dfrac{\prod_{k=1}^n(k+p)}{n^p\prod_{k=1}^nk}\\
&=\dfrac{\prod_{k=p+1}^{n+p}k}{n^p\prod_{k=1}^nk}\\
&=\dfrac{\prod_{k=p+1}^{n}k\prod_{k=n+1}^{n+p}k}{n^p\prod_{k=1}^pk\prod_{k=p+1}^nk}
\qquad\text{assuming } n > p\\
&=\dfrac{\prod_{k=n+1}^{n+p}k}{n^p\prod_{k=1}^pk}\\
&=\dfrac{\prod_{k=1}^{p}(n+k)}{n^pp!}\\
&=\dfrac{\prod_{k=1}^{p}(1+k/n)}{p!}\\
&>\dfrac{1}{p!}\\
\text{and}\\
L(p, n)
&<\dfrac{(1+p/n)^p}{p!}\\
&<\dfrac{(1+1/n^{2/3})^{n^{1/3}}}{p!}
\qquad\text{if } n > p^3\\
&<\dfrac{e^{1/n^{1/3}}}{p!}
\qquad\text{since } (1+1/m^2)^m < e^{1/m}\\
&\to\dfrac{1}{p!}
\qquad\text{as } n \to \infty\\
\end{array}
$