If $\sum\limits_{n=1}^\infty a_n$ is convergent, then so is $\sum\limits_{n=1}^\infty\frac{a_n}{n}$ If $\sum\limits_{n=1}^\infty a_n$ converges, is the series $\sum\limits_{n=1}^\infty\dfrac{a_n}{n}$ also convergent? 
I've found an answer online that said the second series is convergent,but it did't give the proof,I have no idea how to prove it is true or not.
 A: Set
$$
s_n=a_1+\cdots+a_n, \quad n\in\mathbb N.
$$
Then $\{s_n\}$ converges, say to $s\in\mathbb R$. Next
$$
\sum_{k=1}^n\frac{a_k}{k}=\sum_{k=1}^n\frac{s_k-s_{k-1}}{k}=
\sum_{k=1}^n\frac{s_k}{k}-
\sum_{k=2}^n\frac{s_{k-1}}{k}=
\sum_{k=1}^n\frac{s_k}{k}-
\sum_{k=1}^{n-1}\frac{s_{k}}{k+1}\\=
\sum_{k=2}^ns_k\left(\frac{1}{k}-\frac{1}{k+1}\right)+\frac{s_n}{n}
=
\sum_{k=2}^n\frac{s_k}{k(k+1)}+\frac{s_n}{n}.
$$ 
Clearly, $\dfrac{s_n}{n}\to 0$, and 
$$
\sum_{k=2}^n\frac{|s_k|}{k(k+1)}\le M\sum_{k=2}^n\frac{1}{k(k+1)}< \frac{M}{2},
$$
where $M$ is an upper bound of $\{|s_n|\}$, and hence
$$
\sum_{k=2}^n\frac{s_k}{k(k+1)}
$$
converges, due to the Comparison Test.
A: Dirichlet's test states that if the partial sums of $\{a_n\}_{n\geq 1}$ are bounded and $\{b_n\}_{n\geq 1}$ is decreasing towards zero then $\sum_{n\geq 1}a_n b_n$ is convergent (by summation by parts). If $\sum_{n\geq 1}a_n$ is convergent its partial sums are clearly bounded, and $\left\{\frac{1}{n}\right\}_{n\geq 1}$ is blatantly decreasing towards zero. It follows that
$$ \sum_{n\geq 1}a_n\text{ is convergent}\quad\Longrightarrow\quad \sum_{n\geq 1}\frac{a_n}{n}\text{ is convergent}.$$
Kronecker's lemma provides a sort of converse statement:
$$ \sum_{n\geq 1}\frac{a_n}{n}\text{ is convergent}\quad\Longrightarrow\quad a_1+a_2+\ldots+a_n = o(n).$$
