Calculate the following series using telescoping I have the following series
$$\sum_{k=1}^\infty \frac{(k-1)!}{(k+N)!},\quad\text{where }N \in \mathbb{N}. $$
I have found out that the series is equal to
$$\sum_{k=1}^\infty \biggl(\frac{1}{k}\cdot \frac{1}{k+1}\cdot _{...} \cdot \frac{1}{k+N}\biggr)$$
I also know that for $N=1$ we can use partial fraction expansion and we get a telescoping sum of $1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + ... + \frac{1}{k+1}$ which leaves $1$ if $k$ is heading to $\infty$.
Does anyone have any idea how I can go about doing this ? 
 A: To avoid a division by $0$, you should really start at $k=1$. Using @DanielFischer's hint, the telescoping goes as$$\begin{align}\sum_{k\ge1}\frac{1}{k\cdots(k+N)}&=\frac1N\sum_{k\ge1}\frac{k+N-k}{k\cdots(k+N)}\\&=\frac1N\sum_{k\ge1}\left(\frac{k+N}{k\cdots(k+N)}-\frac{k}{k\cdots(k+N)}\right)\\&=\frac1N\sum_{k\ge1}\left(\frac{1}{k\cdots(k+N-1)}-\frac{1}{(k+1)\cdots(k+N)}\right)\\&=\frac1N\left(\left.\frac{1}{k\cdots(k+N-1)}\right|_{k=1}-\lim_{k\to\infty}\frac{1}{k\cdots(k+N-1)}\right)\\&=\frac1N\left(\frac{1}{N!}-0\right)\\&=\frac{1}{N\cdot N!}.\end{align}$$
A: The neat thing about
$p_n(x)
=\prod_{k=0}^{n-1}(x+k)
$
is that is telescopes in both
numerator and denominator.
$\begin{array}\\
p_n(x+1)-p_n(x)
&=\prod_{k=0}^{n-1}(x+1+k)-\prod_{k=0}^{n-1}(x+k)\\
&=\prod_{k=1}^{n}(x+k)-\prod_{k=0}^{n-1}(x+k)\\
&=(x+n)\prod_{k=1}^{n-1}(x+k)-x\prod_{k=1}^{n-1}(x+k)\\
&=((x+n)-x)\prod_{k=1}^{n-1}(x+k)\\
&=n\prod_{k=0}^{n-2}(x+1+k)\\
&=np_{n-1}(x+1)\\
\text{so}\\
\dfrac{p_n(x+1)-p_n(x)}{n}
&=p_{n-1}(x+1)\\
\text{or}\\
\dfrac{p_{n+1}(x)-p_{n+1}(x-1)}{n+1}
&=p_{n}(x)\\
\text{and}\\
\dfrac1{p_n(x)}-\dfrac1{p_n(x+1)}
&=\dfrac1{\prod_{k=0}^{n-1}(x+k)}-\dfrac1{\prod_{k=0}^{n-1}(x+1+k)}\\
&=\dfrac1{\prod_{k=0}^{n-1}(x+k)}-\dfrac1{\prod_{k=1}^{n}(x+k)}\\
&=\dfrac{x+n}{\prod_{k=0}^{n}(x+k)}-\dfrac{x}{\prod_{k=0}^{n}(x+k)}\\
&=\dfrac{(x+n)-x}{\prod_{k=0}^{n}(x+k)}\\
&=\dfrac{n}{p_{n+1}(x)}\\
\text{so}\\
\dfrac1{n}\left(\dfrac1{p_n(x)}-\dfrac1{p_n(x+1)}\right)
&=\dfrac{1}{p_{n+1}(x)}\\
\text{or}\\
\dfrac1{n-1}\left(\dfrac1{p_{n-1}(x)}-\dfrac1{p_{n-1}(x+1)}\right)
&=\dfrac{1}{p_{n}(x)}\\
\end{array}
$
