Show that $\sum\limits_{k=1}^\infty \frac{1}{k(k+1)(k+2)\cdots (k+p)}=\frac{1}{p!p} $ for every positive integer $p$ I have to prove that 
$$\sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)\cdots (k+p)}=\dfrac{1}{p!p}$$
How can I do that?
 A: Simpler answer: Telescope
for fixed $p\in\Bbb N$  If we let 
$u_n =\frac1{n(n+1)\cdots(n+p-1)}$
then, 
$u_n -u_{n+1} = \frac1{n(n+1)\cdots(n+p-1)}-\frac1{(n+1)(n+2)\cdots(n+p)}
=\frac p{n(n+1)\cdots(n+p)}$
Hence By Telescoping sum we get, $$ u_1 = u_1-\lim_{n\to \infty}u_{n+1}=\sum_{n=1}^{\infty} (u_n -u_{n+1}) = p\sum_{n=1}^{\infty}  \frac{1}{n(n+1)(n+2)\cdots (n+p)}$$
But $u_1 = \frac1{1(1+1)\cdots(1+p-1)} = \frac{1}{p!}$
Hence $$ \color{red}{\sum_{n=1}^{\infty}  \frac{1}{n(n+1)(n+2)\cdots (n+p)} = \frac{1}{p!p}}$$
A: Note that
$$
\frac{1}{k(k\!+\!1)(k\!+\!2)\cdots (k\!+\!p)}=\frac{1}{p}\!\left(\frac{1}{k(k\!+\!1)\cdots (k\!+\!p\!-\!1)}-\frac{1}{(k\!+\!1)(k\!+\!2)\cdots (k\!+\!p)}\right)
$$
Hence
$$
\sum_{k=1}^n\frac{1}{k(k+1)(k+2)\cdots (k+p)}=\\=\frac{1}{p}\left(\frac{1}{p!}-\frac{1}{(n+1)(n+2)(n+3)\cdots (n+p)}\right) \to \frac{1}{p!p}
$$
as $n\to\infty$.
Similarly, 
$$
\sum_{k=1}^n\frac{1}{(k+1)(k+2)\cdots (k+p)}=\\=\frac{1}{p-1}\left(\frac{1}{p!}--\frac{1}{(n+2)(n+2)(n+3)\cdots (n+p)}\right) \to \frac{1}{p!(p-1)}
$$
