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.
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.
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: $$ \sum_{n\geq 1}\frac{a_n}{n}\text{ is convergent}\quad\Longrightarrow\quad a_1+a_2+\ldots+a_n = o(n).$$
