If $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$

Let $a_1,a_2,a_3,\ldots$ be reals. Prove that if $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$.

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.

Solution 1:

What you did is correct; in fact you can show that if $\{a_n\}$ and $\{b_n\}$ are two sequences of real numbers and $\sum_{n\geq 0}a_n^2$ and $\sum_{n\geq 0}b_n^2$ are convergent then the series $\sum_{n=0}^{+\infty}|a_nb_n|$ is convergent, noting that $0\leq |a_nb_n|\leq \max(a_n^2,b_n^2)\leq a_n^2+b_n^2$.

Your particular case is $b_n=\frac 1n$.

Solution 2:

Another approach is to note that for any positive integer $N,$ we have $\sum_{n=1}^{N} \frac{|a_n|}{n} \leq \sqrt{ \sum_{n= 1}^{N} a_{n}^{2}} \sqrt{ \sum_{n=1}^{N} \frac{1}{n^2}}$, and this is in turn less than $\frac{\pi}{\sqrt{6}}\sqrt{ \sum_{n= 1}^{N} a_{n}^{2} }.$ The first inequality follows by the Cauchy-Schwarz inequality, and the second follows by Euler's formula $\frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2}.$