December 17th, 2018, 03:51 PM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68  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: 578 Thanks: 345 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: 7,618 Thanks: 2608 Math Focus: Mainly analysis and algebra 
For positive ${x_n}$.

December 18th, 2018, 04:48 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
$\displaystyle \left(\leftx_n\right  \frac1n\right)^2 \geqslant 0 \implies \left\frac{2x_n}{n}\right \leqslant x_n^2 + \frac{1}{n^2} \implies \sum_{n=1}^\infty \frac{x_n}{n}$ is absol.utely convergent.


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