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

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?

• (in particular, through the fact that $\sum_{n\in\Bbb N} \frac{\lvert a_n\rvert}n$ converges). – user228113 Feb 23 '17 at 12:34
• math.stackexchange.com/questions/112579/… – StubbornAtom Dec 6 '19 at 21:27

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