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 ?

Solution 1:

To avoid a division by $0$, you should really start at $k=1$. Using @DanielFischer's hint, the telescoping goes as$$\begin{align}\sum_{k\ge1}\frac{1}{k\cdots(k+N)}&=\frac1N\sum_{k\ge1}\frac{k+N-k}{k\cdots(k+N)}\\&=\frac1N\sum_{k\ge1}\left(\frac{k+N}{k\cdots(k+N)}-\frac{k}{k\cdots(k+N)}\right)\\&=\frac1N\sum_{k\ge1}\left(\frac{1}{k\cdots(k+N-1)}-\frac{1}{(k+1)\cdots(k+N)}\right)\\&=\frac1N\left(\left.\frac{1}{k\cdots(k+N-1)}\right|_{k=1}-\lim_{k\to\infty}\frac{1}{k\cdots(k+N-1)}\right)\\&=\frac1N\left(\frac{1}{N!}-0\right)\\&=\frac{1}{N\cdot N!}.\end{align}$$

Solution 2:

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} $