
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
May 10th, 2015, 10:15 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, 10:32 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,035 Thanks: 2343 Math Focus: Mainly analysis and algebra 
The partial fraction forms a telescoping sum. I imagine you can prove this by induction.

May 10th, 2015, 02:23 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,379 Thanks: 542 
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, 02:30 PM  #4 
Member Joined: Mar 2015 From: Brasil Posts: 90 Thanks: 4  
May 10th, 2015, 05:51 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,642 Thanks: 960 Math Focus: Elementary mathematics and beyond 
$\displaystyle \sum_{n=1}^\infty\frac{1}{n(n+1)...(n+p)}=\frac{1} {p\cdot p!}$


Tags 
convergence, series, sum 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Sum of series and convergence  Luiz  Real Analysis  3  March 26th, 2015 09:08 AM 
Sum of series and convergence  Luiz  Real Analysis  4  March 25th, 2015 02:17 PM 
Sum of series and convergence  Luiz  Real Analysis  1  March 22nd, 2015 02:22 PM 
convergence of series  mghg13  Calculus  3  February 17th, 2013 02:26 AM 
Another Series Convergence  HairOnABiscuit  Real Analysis  1  April 28th, 2010 12:41 AM 