# 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?

#### Solutions Collecting From Web of "General formula for this sum $\sum_{k=1}^n\frac{1}{k(k+1)…(k+m)}$"

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

This is a telescoping sum. The result then is

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