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I have a series question which I can not solve for two days.

$$\sum_{n=1}^∞\frac{1}{(n+1)(n+2)(n+3)....(n+p)} = \frac{1}{p.p!}$$

How can I prove and solve this equation?

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    Write numerator as $$\dfrac{n+p-(n+1)}{p-1}$$ and use Telescoping Series – 2017-02-09
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    Your formula would be correct if your denominator was $n(n+1)\cdots(n+p)$. – 2017-02-09
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    You may use creative telescoping or Euler's Beta function as shown here: http://math.stackexchange.com/questions/1964430/funny-double-infinite-sum/1964478#1964478 – 2017-02-09
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    This has been asked at least a couple of times before, I don't know the links off hand but it will likely be closed as duplicate when someone gets around to finding them (Jack's is close but asks for more). – 2017-02-09

2 Answers 2

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Note the following $$\sum_{n=1}^∞\frac{1}{(n+1)(n+2)(n+3)....(n+p)}$$

Using partial fraction decomposition the summation becomes $$\frac{1}{p-1}\sum_{n=1}^∞ \left(\frac{1}{(n+1)(n+2)....(n+p-1)}-\frac{1}{(n+2)(n+3)....(n+p)}\right)$$ We can use the telescoping series. The consecutive terms cancel. So your answer is incorrect. The correct answer would be $$\frac{1}{p-1} \times \frac{1}{(1+1)(1+2)....(1+p-1)}=\frac{1}{(p-1)p!}$$ We are done! As mentioned by @ThomasAndrews, the formula would be correct if the denominator was $n(n+1)(n+2)(n+3)....(n+p)$ which can be confirmed through the same method.

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    How did we find $$ \frac{1}{p-1}$$ in the second line? – 2017-02-09
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    @Onur $\frac{1}{(n+1)(n+2)(n+3)....(n+p)}$ is $$\frac{1}{p-1} \times \frac{(n+p)-(n+1)}{(n+1)(n+2)(n+3)....(n+p)}$$Which becomes $$\frac{1}{p-1} \times \frac{1}{(n+1)(n+2)....(n+p-1)}-\frac{1}{(n+2)(n+3)....(n+p)}$$ – 2017-02-09
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    @Onur Do you understand? – 2017-02-09
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    Yes, now I certainly saw it. Thanks for the help. I would call the publisher to warn them about their mistake. :) – 2017-02-09
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    @Onur You get money if you do that in my country :) I hope you would accept my answer? – 2017-02-09
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    @ S.C.B Of course. Your answer was the exact one I was looking for. I hope they give money here as well, though I doubt it. :) – 2017-02-09
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    @ S.C.B I already upvoted. But it seems my reputation is not high enough for my vote to be recorded. – 2017-02-09
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    @Onur You need 15 reputation for that privillege, unfortuantely. – 2017-02-09
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    @Onur Though you can still accept (i.e., click on the check button button) at that level. – 2017-02-09
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$$\frac{1}{(n+1)(n+2)(n+3)....(n+p)} = \frac{n!}{(n+p)!}=\frac{1}{\Gamma(p)}\frac{\Gamma(n+1)\Gamma(p)}{\Gamma(n+p+1)}=\frac{1}{\Gamma(p)}\beta(n+1,p)$$ then \begin{eqnarray} \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)(n+3)....(n+p)} &=& \sum_{n=1}^\infty\frac{1}{\Gamma(p)}\beta(n+1,p)\\ &=& \frac{1}{\Gamma(p)}\sum_{n=1}^\infty\int_0^1t^{n}(1-t)^{p-1}dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1\sum_{n=1}^\infty t^{n}(1-t)^{p-1}dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1\frac{t}{1-t}(1-t)^{p-1}dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1t(1-t)^{p-2}dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1t^{p-2}(1-t)dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1t^{p-2}-t^{p-1}dt\\ &=& \frac{1}{\Gamma(p)}\frac{1}{p(p-1)}\\ &=& \frac{1}{p!(p-1)} \end{eqnarray} For $p=2$ we know $$\sum_{n=1}^\infty\frac{1}{(n+1)(n+2)}=\sum_{n=1}^\infty\frac{1}{n+1}-\frac{1}{n+2}=\frac12$$ that shows our answer is true.