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 May 10th, 2015, 09:15 AM #1 Member   Joined: Mar 2015 From: Brasil Posts: 90 Thanks: 4 Convergence and Series Sum For all fixed $p\in \mathbb{N}$, prove that the series $\sum _{1}^{\infty }\frac{1}{n(n+1)...(n+p)}$ It is convergent and compute their sum.
 May 10th, 2015, 09:32 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra The partial fraction forms a telescoping sum. I imagine you can prove this by induction.
 May 10th, 2015, 01:23 PM #3 Global Moderator   Joined: May 2007 Posts: 6,527 Thanks: 588 As long as p > 0, $\displaystyle \frac{1}{n(n+1)...(n+p)} < \frac{1}{n^2}$. The sum over the latter converges.
May 10th, 2015, 01:30 PM   #4
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Quote:
 Originally Posted by mathman As long as p > 0, $\displaystyle \frac{1}{n(n+1)...(n+p)} < \frac{1}{n^2}$. The sum over the latter converges.

And the sum how do I calculate ?????

 May 10th, 2015, 04:51 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond $\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+1)...(n+p)}=\frac{1} {p\cdot p!}$

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