My Math Forum Convergence and Series Sum

 Real Analysis Real Analysis Math Forum

 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: 6,775 Thanks: 2194 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: 6,234 Thanks: 496 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

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: 7,488 Thanks: 887 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 Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Luiz Real Analysis 3 March 26th, 2015 08:08 AM Luiz Real Analysis 4 March 25th, 2015 01:17 PM Luiz Real Analysis 1 March 22nd, 2015 01:22 PM mghg13 Calculus 3 February 17th, 2013 02:26 AM HairOnABiscuit Real Analysis 1 April 27th, 2010 11:41 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top