# 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 ?

• Start by looking at $\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}$. If that's not enough, look at $\frac{1}{k(k+1)(k+2)} - \frac{1}{(k+1)(k+2)(k+3)}$ too. – Daniel Fischer Dec 4 '19 at 22:10
• Sorry but that really didn't help. I'm still going in circles. – Zntzozt Dec 4 '19 at 22:33
• – robjohn Dec 4 '19 at 23:44

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}
• This is brilliant. But can you please explain how you got $N!$ at the end. – Zntzozt Dec 4 '19 at 22:38
The neat thing about $$p_n(x) =\prod_{k=0}^{n-1}(x+k)$$ is that is telescopes in both numerator and denominator.
$$\begin{array}\\ p_n(x+1)-p_n(x) &=\prod_{k=0}^{n-1}(x+1+k)-\prod_{k=0}^{n-1}(x+k)\\ &=\prod_{k=1}^{n}(x+k)-\prod_{k=0}^{n-1}(x+k)\\ &=(x+n)\prod_{k=1}^{n-1}(x+k)-x\prod_{k=1}^{n-1}(x+k)\\ &=((x+n)-x)\prod_{k=1}^{n-1}(x+k)\\ &=n\prod_{k=0}^{n-2}(x+1+k)\\ &=np_{n-1}(x+1)\\ \text{so}\\ \dfrac{p_n(x+1)-p_n(x)}{n} &=p_{n-1}(x+1)\\ \text{or}\\ \dfrac{p_{n+1}(x)-p_{n+1}(x-1)}{n+1} &=p_{n}(x)\\ \text{and}\\ \dfrac1{p_n(x)}-\dfrac1{p_n(x+1)} &=\dfrac1{\prod_{k=0}^{n-1}(x+k)}-\dfrac1{\prod_{k=0}^{n-1}(x+1+k)}\\ &=\dfrac1{\prod_{k=0}^{n-1}(x+k)}-\dfrac1{\prod_{k=1}^{n}(x+k)}\\ &=\dfrac{x+n}{\prod_{k=0}^{n}(x+k)}-\dfrac{x}{\prod_{k=0}^{n}(x+k)}\\ &=\dfrac{(x+n)-x}{\prod_{k=0}^{n}(x+k)}\\ &=\dfrac{n}{p_{n+1}(x)}\\ \text{so}\\ \dfrac1{n}\left(\dfrac1{p_n(x)}-\dfrac1{p_n(x+1)}\right) &=\dfrac{1}{p_{n+1}(x)}\\ \text{or}\\ \dfrac1{n-1}\left(\dfrac1{p_{n-1}(x)}-\dfrac1{p_{n-1}(x+1)}\right) &=\dfrac{1}{p_{n}(x)}\\ \end{array}$$