Assume that a series $\sum\limits_{n\geqslant1} a_n$ is convergent. Does this imply that $\sum\limits_{n\geqslant1}\frac{a_n}{n}$ is absolutely convergent?
My first thought is no, but I'm having a hard time finding a counterexample. If there is a counterexample then it must be conditionally convergent, because if $\sum_n a_n$ is absolutely convergent then $\sum_n\frac{a_n}{n}$ clearly will be as well. The problem is I don't really know many examples of conditionally convergent series.
Of course there's $\sum_n\frac{(-1)^n}{n}$, but that doesn't work. We can extend this to a class of conditionally convergent series $\sum_n\frac{(-1)^n}{n^p}$ for any $0<p\le1$, but none of these work either because we will simply get $\sum_n\frac{(-1)^n}{n^{p+1}}$, which will converge absolutely because $p+1>1$.
So I'm feeling pretty stuck. Anybody have any good suggestions of conditionally convergent series for me to try? Or if my intuition is wrong and this happens to be true, can someone give me a hint on how I'd go about proving it? I'd prefer hints/pushes in the right direction rather than just giving me the answer.
Hint: try $a_n=(-1)^n/\log n$ $(n\geq 2)$. More generally, you can build as many conditionally convergent series as you want using the Alternating series test.