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> Let $a\_1,a\_2,a\_3,\ldots$ be reals. Prove that if $\sum\_{n=1}^{\infty} a\_n^{2}$ converges, then so does $\sum\_{n=1}^{\infty} \frac {a\_n}{n}$.
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For this I have shown the case for when $ a\_n^{2} \le\frac {|a\_n|}{n}$ $\Rightarrow$ $ |a\_n|\le\frac {1}{n}$ $\Rightarrow$ $\frac {|a\_n|}{n} \le \frac{1}{n^{2}}$ and we know that $\sum\_{n=1}^{\infty} \frac {1}{n^{2}}$ converges and hence $\sum\_{n=1}^{\infty}\frac {a\_n}{n}$ converges by the comparison test.
Now considering $ a\_n^{2} \ge\frac {|a\_n|}{n}$ $\Rightarrow$ $\frac {|a\_n|}{n} \le a\_n^{2}$
$\rightarrow$ combining the two cases for any n we have: $\frac {|a\_n|}{n}\le\frac{1}{n^{2}}+a\_n^{2}$ Hence using the comparion test again we know that $\sum\_{n=1}^{\infty} a\_n^{2}$ converges and $\sum\_{n=1}^{\infty} \frac {1}{n^{2}}$ converges hence the sum converges so we can conclude that $\sum\_{n=1}^{\infty} \frac {a\_n}{n}$ is absoluetly convergent $\Rightarrow$ convergent. 
Not to sure if this is correct, any help would be much appreciated, many thanks.