Can I conclude from $\sum_{n=1}^\infty u_n$ being convergent that $\sum_{n=1}^\infty (-1)^n\frac{u_n}{n}$ is convergent?

Question

So the question is the following statement correct? $$ \sum_{n=1}^\infty u_n < \infty \implies \sum_{n=1}^\infty (-1)^n\frac{u_n}{n} < \infty. $$ I think the answer is not, however I cannot find a counterexample or prove it otherwise.

Answer 1

Hint: What if $u_n=\frac{(-1)^n}{\log(n+1)}$?

Answer 2

It's not true generally, but perhaps could be made true under certain conditions. For example, if all $u_n > 0$, then $$ (-1)^n \frac{u_n}{n}<\frac{u_n}{n} < u_n, \quad \forall n\ge 2, $$ so your statement would hold.

Additionally, this suggests a good place to start looking for counter-examples if you want to claim it always holds.