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 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 $\frac{2x_n}{n} \leq x_n^2 + \frac{1}{n^2}$. Thanks from topsquark, v8archie and idontknow
 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 $\displaystyle \left(\left|x_n\right| - \frac1n\right)^2 \geqslant 0 \implies \left|\frac{2x_n}{n}\right| \leqslant x_n^2 + \frac{1}{n^2} \implies \sum_{n=1}^\infty \frac{x_n}{n}$ is absol.utely convergent. Thanks from idontknow

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