# General formula for this sum $\sum_{k=1}^n\frac{1}{k(k+1)...(k+m)}$

Is there a general formula for

$\sum_{k=1}^n\frac{1}{k(k+1)...(k+m)}$?

I know that the limit is $\frac{1}{mm!}$ but is there a combinatorial expression for this?

• Your right. $k$ has to start with $1$. Edited. Commented Nov 7, 2016 at 12:54
• are you looking for an algebraic formula or for a combinatorial argumentation of that? Commented Nov 7, 2016 at 12:56
• In fact I found something in between: $\frac{1}{a}(\frac{1}{1^{\bar{a}}}-\frac{1}{(n+1)^{\bar{a}}})$, where $n^{\bar{a}}=n(n+1)...(n+a-1)$. And its works! Is there another combinatorial? Commented Nov 7, 2016 at 13:04
• If you're asking for an expression of the sum: $$\sum_{k=1}^n \frac{1}{k(k+1)\cdots(k+m)} = \sum_{k=1}^n \frac{(k-1)!}{(k+m)!} = \frac{1}{mm!} - \frac{n!}{m(m+n)!}$$ Commented Nov 7, 2016 at 13:08
• Thank you. This is the same i found. Is there a combinatorial as well? Commented Nov 7, 2016 at 13:11

$$\sum_{k=1}^n \frac{1}{k(k+1)\cdots(k+m)} = \frac1m\sum_{k=0}^n \frac{(k + m) - k}{k(k+1)\cdots(k+m)} \\ = \frac1m\sum_{k=1}^n \left[ \frac{1}{k(k+1)\cdots(k+m-1)} - \frac{1}{(k+1)(k+2)\cdots(k+m)}\right]$$
$$-\frac1m\left[ \frac{1}{(n+1)(n+2)\cdots(n+m)} - \frac{1}{1\cdot 2\cdots m}\right ] \\ = -\frac1m\left[ \frac{n!}{(m+n)!} - \frac{1}{m!}\right] \\ = \frac{1}{mm!} - \frac{n!}{m(m+n)!}$$