Find the sum $$\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}$$
My idea: since $$\dfrac{1}{n(n+1)(n+2)\cdots(n+2014)}=\dfrac{1}{2014}\left(\dfrac{1}{n(n+1)(n+2)\cdots(n+2013)}-\dfrac{1}{(n+1)(n+2)\cdots(n+2014)}\right)?$$
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Where is this problem from? what have you tried? – ShreevatsaR Mar 05 '14 at 12:34
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Taking logarithm and then integral...doesn't that help? – Hawk Mar 05 '14 at 12:34
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2See http://math.stackexchange.com/questions/477174/a-convergent-series – user91500 Mar 05 '14 at 12:41
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Your idea should work, giving a telescoping series. Sneaky! – vonbrand Mar 05 '14 at 13:24
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Hint
The telescoping sum is useful $$\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}=\frac1{2014}\dfrac{(n+2014)-n}{n(n+1)(n+2)(n+3)\cdots(n+2014)}=\frac 1{2014}\left(\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+2013)}-\dfrac{1}{(n+1)(n+2)(n+3)\cdots(n+2014)}\right)$$
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1@Hawk: See the Wikipedia article on telescoping series, which introduces an example very similar to this one. – ShreevatsaR Mar 05 '14 at 12:41
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@ShreevatsaR Ah yes, I just noticed that it has a summation symbol, I was thinking it to be a limit symbol. – Hawk Mar 05 '14 at 12:42
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Just by curiosity, I used a CAS for the computation of $$S(m)=\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+m)}$$ The first result was quite amazing $$S(m)=\frac{\Gamma (m) \Gamma (m+2)}{(2)_m \Gamma (m+1)^2}$$ where appear Pochhammer numbers. Fortunately, this simplifies to $$S(m)=\frac{1}{m^2 \Gamma (m)}$$ which is a quite nice closed form.
Claude Leibovici
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