Real Analysis Real Analysis Math Forum

 December 17th, 2018, 04:51 PM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 828 Thanks: 113 Math Focus: Elementary Math 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 05:49 AM. December 17th, 2018, 08:26 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 684 Thanks: 459 Math Focus: Dynamical systems, analytic function theory, numerics For any $n$, you have $(x_n - \frac{1}{n})^2 > 0$ which implies the bound $\frac{2x_n}{n} \leq x_n^2 + \frac{1}{n^2}$. Thanks from topsquark, v8archie and idontknow December 18th, 2018, 04:33 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra For positive ${x_n}$. December 18th, 2018, 05:48 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 $\displaystyle \left(\left|x_n\right| - \frac1n\right)^2 \geqslant 0 \implies \left|\frac{2x_n}{n}\right| \leqslant x_n^2 + \frac{1}{n^2} \implies \sum_{n=1}^\infty \frac{x_n}{n}$ is absol.utely convergent. Thanks from idontknow Tags convergence, proof Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post v8archie Real Analysis 1 December 24th, 2013 12:51 PM MageKnight Real Analysis 1 May 2nd, 2013 07:40 PM William_33 Applied Math 1 March 4th, 2013 04:19 PM jmu123 Real Analysis 3 December 6th, 2010 02:35 PM zve5 Real Analysis 2 September 24th, 2008 03:32 PM

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