# Find the sum of the series $\sum^{\infty}_{n=1} \frac{1}{(n+1)(n+2)(n+3) \cdots (n+k)}$

Find the sum of the series

$$\sum^{\infty}_{n=1} \frac{1}{(n+1)(n+2)(n+3) \cdots (n+k)}$$

Given series

$$\sum^{\infty}_{n=1} \frac{1}{(n+1)(n+2)(n+3) \cdots (n+k)}$$

$$= \frac{1}{2\cdot3\cdot4 \cdots (k+1)}+\frac{1}{3\cdot4\cdot5 \cdots (k+2)}+\frac{1}{4\cdot5\cdot6\cdots (k+3)} +\cdots$$

now how to proceed further in this pleas suggest thanks ....

• $$= \frac{1}{2\cdot3\cdot4 \cdots (k+1)}+\frac{1}{3\cdot4\cdot5 \cdots (k+2)}+\frac{1}{4\cdot5\cdot6\cdots (k+3)} + \frac{1}{5\cdot6\cdot7\cdots (k+4)} +\cdots$$ – Jihad Jan 11 '15 at 16:04
• $$\frac{1}{(n+1)(n+2)(n+3) \cdots (n+k)} = \frac{1}{k-1}\left(\frac{(n+k)-(n+1)}{(n+1)(n+2)(n+3) \cdots (n+k)}\right)\\ = \frac{1}{k-1}\left(\frac{1}{(n+1)(n+2) \cdots (n+k-1)} - \frac{1}{(n+2)(n+3) \cdots (n+k)}\right)$$ – achille hui Jan 11 '15 at 16:07

For $k>1$,
\begin{align} S &=\sum_{n=1}^{\infty}\frac{1}{(n+1)(n+2)(n+3)\dots(n+k)}\\ &=\sum_{n=1}^{\infty}\frac{n!}{(n+k)!}\\ &=\frac{1}{(k-1)!}\sum_{n=1}^{\infty}\frac{n!(k-1)!}{(n+k)!}\\ &=\frac{1}{(k-1)!}\sum_{n=1}^{\infty}\frac{\Gamma{(n+1)}\,\Gamma{(k)}}{\Gamma{(n+k+1)}}\\ &=\frac{1}{(k-1)!}\sum_{n=1}^{\infty}\operatorname{B}{\left(n+1,k\right)}\\ &=\frac{1}{(k-1)!}\sum_{n=1}^{\infty}\int_{0}^{1}t^n(1-t)^{k-1}\,\mathrm{d}t\\ &=\frac{1}{(k-1)!}\int_{0}^{1}(1-t)^{k-1}\sum_{n=1}^{\infty}t^n\,\mathrm{d}t\\ &=\frac{1}{(k-1)!}\int_{0}^{1}(1-t)^{k-1}\frac{t}{1-t}\,\mathrm{d}t\\ &=\frac{1}{(k-1)!}\int_{0}^{1}t\,(1-t)^{k-2}\,\mathrm{d}t\\ &=\frac{1}{(k-1)!}\operatorname{B}{\left(2,k-1\right)}\\ &=\frac{1}{(k-1)!}\cdot\frac{1}{k(k-1)}\\ &=\frac{1}{k!(k-1)}.\\ \end{align}