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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: 7,276 Thanks: 2437 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: 6,497 Thanks: 580 
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 
Member Joined: Mar 2015 From: Brasil Posts: 90 Thanks: 4  
May 10th, 2015, 04:51 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,788 Thanks: 1037 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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convergence, series, sum 
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