December 17th, 2018, 03:51 PM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 646 Thanks: 92  Convergence proof
How to show that if $\displaystyle \sum_{n=1}^\infty x^2_n $ converges, then $\displaystyle \sum_{n=1}^\infty \frac{x_n}{n} $ also converges?
Last edited by skipjack; December 18th, 2018 at 04:49 AM. 
December 17th, 2018, 07:26 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 646 Thanks: 411 Math Focus: Dynamical systems, analytic function theory, numerics 
For any $n$, you have $(x_n  \frac{1}{n})^2 > 0$ which implies the bound $\frac{2x_n}{n} \leq x_n^2 + \frac{1}{n^2}$.

December 18th, 2018, 03:33 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra 
For positive ${x_n}$.

December 18th, 2018, 04:48 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,975 Thanks: 2225 
$\displaystyle \left(\leftx_n\right  \frac1n\right)^2 \geqslant 0 \implies \left\frac{2x_n}{n}\right \leqslant x_n^2 + \frac{1}{n^2} \implies \sum_{n=1}^\infty \frac{x_n}{n}$ is absol.utely convergent.


Tags 
convergence, proof 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Proof of Convergence  v8archie  Real Analysis  1  December 24th, 2013 11:51 AM 
Proof of Convergence of a (recursive) sequence  MageKnight  Real Analysis  1  May 2nd, 2013 06:40 PM 
Convergence proof  William_33  Applied Math  1  March 4th, 2013 03:19 PM 
proof of convergence....please help!  jmu123  Real Analysis  3  December 6th, 2010 01:35 PM 
Sequence convergence proof  zve5  Real Analysis  2  September 24th, 2008 02:32 PM 