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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 ....

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    $\begingroup$ $$ = \frac{1}{2\cdot3\cdot4 \cdots (k+1)}+\frac{1}{3\cdot4\cdot5 \cdots (k+2)}+\frac{1}{4\cdot5\cdot6\cdots (k+3)} + \frac{1}{5\cdot6\cdot7\cdots (k+4)} +\cdots$$ $\endgroup$ – Jihad Jan 11 '15 at 16:04
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    $\begingroup$ $$\frac{1}{(n+1)(n+2)(n+3) \cdots (n+k)} = \frac{1}{k-1}\left(\frac{(n+k)-(n+1)}{(n+1)(n+2)(n+3) \cdots (n+k)}\right)\\ = \frac{1}{k-1}\left(\frac{1}{(n+1)(n+2) \cdots (n+k-1)} - \frac{1}{(n+2)(n+3) \cdots (n+k)}\right)$$ $\endgroup$ – achille hui Jan 11 '15 at 16:07
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There is a technique involving the machinery of beta functions which comes close to providing an algorithmic approach to evaluating a broad family of sums like these involving ratios of factorials. The idea to is to represent the ratio of factorials as a beta function, write the beta function as an integral, move the summation under the integral sign, and then evaluate the resulting geometric series. The whole process transforms the series into a integral.

For $k>1$,

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

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  • $\begingroup$ Very nice solution. (+1)! $\endgroup$ – user98186 Jan 29 '16 at 17:20
  • $\begingroup$ Very Nice solution. $\endgroup$ – juantheron Apr 12 '16 at 3:54

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