$\sum_{n=1}^{\infty}a_n$ 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?

 A: I wrote in a comment that this proof is correct.
It should be added that the stronger result holds :
Proposition If $\sum_{n\ge1}a_n$ converges, where the $a_n$ are complex, and if $\alpha$ is some positive constant, then $\sum_{n\ge1}\frac{a_n}{n^\alpha}$ converges to.
This is a consequence of the following theorem :
Theorem (Abel) If $(a_n)\in\mathbb{C}^\mathbb{N}$ and $(t_n)\in[0,+\infty[^\mathbb{N}$ are sequences such that :


*

*$\exists M\ge0\,\forall n\in\mathbb{N},\,\left|\sum_{k=0}^na_k\right|\le M$

*$\forall n\in\mathbb{N},\,t_{n+1}\le t_n$ and $\lim_{n\to\infty}t_n=0$
then the series $\sum_{n\ge0}t_na_n$ converges.
Proof If we define $A_n=\sum_{k=0}^na_k$ for all $n\in\mathbb{N}$, then (Abel's transformation) :
$$\sum_{k=1}^nt_ka_k=\sum_{k=1}^nt_k(A_k-A_{k-1})=-t_1A_0+t_nA_n+\sum_{k=1}^{n-1}(t_t-t_{k+1})A_k$$
The series $\sum_{k\ge1}(t_k-t_{k+1})A_k$ converges absolutely (and hence converges) because $\left|(t_k-t_{k+1})A_k\right|\le M(t_k-t_{k+1})$, for all $k$.
Clearly the sequence $(t_nA_n)_{n\ge0}$ converges (to zero) and this completes the proof.
A: 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.
