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If $\sum a^2_n$ converges so does $\sum \frac{a_n}{n}$

I'm trying to prove/ disprove this statement. But I couldn't end up with anything yet. Any help?

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  • $\begingroup$ (in particular, through the fact that $\sum_{n\in\Bbb N} \frac{\lvert a_n\rvert}n$ converges). $\endgroup$ – user228113 Feb 23 '17 at 12:34
  • $\begingroup$ See also here. $\endgroup$ – user228113 Feb 23 '17 at 12:36
  • $\begingroup$ math.stackexchange.com/questions/112579/… $\endgroup$ – StubbornAtom Dec 6 '19 at 21:27
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Hint: Using the following inequallity $(a_n-\frac{1}{n})^2\ge 0$ one has $$ a^2_n+\frac{1}{n^2}\ge 2 a_n \frac{1}{n}$$. Dividing by 2 , taking sums you have the convergence of the greater sum, so your series converges

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