TITLE: Is there a sequence such that $\sum{a_n}$ diverges but $\sum{na_n}$converges?
QUESTION [11 upvotes]: Is there a (real) sequence such that $\sum{a_n}$ diverges but $\sum{na_n}$converges?

REPLY [15 votes]: No, there is no such sequence.  Suppose that $\displaystyle \sum_{n=1}^\infty na_n$ converges.  Then, since $\{\frac{1}{n}\}_{n\geq 1}$ is monotone and bounded (it goes to zero), we have $\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{n}na_n$ converging by Abel's test (or by Dirichlet's test); see Abel's test  and Dirichlet's test  .

REPLY [7 votes]: The idea is (see applications of Abel transformation on Wikipedia):
If $\sum_n u_n$ is convergent and $v_n \to 0$ monotonically, then $\sum_n u_n v_n$ is convergent.
Here, take $u_n=na_n$ and $v_n=\frac 1 n$.
So the answer is no.