$\sum_{n=1}^{\infty}a_n$ converges, with $a_n >0$ for all $n$, then $\sum_{n=1}^{\infty} \frac{a_n}{n}$ converges I want to prove that if $\sum_{n=1}^{\infty}a_n$ converges, with $a_n >0$ for all $n$, then $\sum_{n=1}^{\infty} \frac{a_n}{n}$ converges.
My book gives a proof where shows that $S_k=\sum_{n=1}^{k} \frac{a_n}{n}$ is monotonically increasing and bounded, so converges by the completeness axiom.
However can I prove it by the comparison test?
I know that $\forall n\in\mathbb{N}$ $$0\leq \frac{a_n}{n} \leq a_n$$ and also, I know that $\sum_{n=1}^{\infty}a_n$ converges, so by the comparison tests, also $\sum_{n=1}^{\infty} \frac{a_n}{n}$ converges.

Is this acceptable?


Since $\sum a_n$ converges, then $a_n\to0$, thus for sufficiently large $n$ we have $0<a_n<1$, multiplying by $a_n$ we get $0<a_n^2<a_n$. Using the comparison test, $\sum a_n^2$ converges. Note that for every $n\in\mathbb{N}$ we have
$$\left(|a_n|-\frac{1}{n}\right)^2\geq0\Longrightarrow 2\frac{|a_n|}{n}\leq a_n^2+\frac{1}{n^2}.$$
Then we can use the comparison test to assert the convergence of $\sum\dfrac{|a_n|}{n}$. Once again, by the same test, we have that $\sum\dfrac{a_n}{n}$ converges.
