"Motivation"/Introduction:
If $(a_n)$ is a decreasing real sequence and $\displaystyle\sum a_n $ converges, then $n a_n \to 0,\ $ for example, by the Cauchy Condensation test.
If $(a_n)$ is a real sequence and $\displaystyle\sum a_n $ converges but $(a_n)$ is not necessarily decreasing, then does $n a_n\to 0?$ No: for example, take:
$ a_n:= \begin{cases} 2^{-n}&\text{if}\, n\neq 2^k\\ \frac{1}{k^2}&\text{if}\, n = 2^k\\ \end{cases} $
Therefore, if $(a_n)$ is a real sequence and $\displaystyle\sum a_n $ converges but $(a_n)$ is not necessarily decreasing, then $\displaystyle\sum n a_n\ $ and $\displaystyle\sum (-1)^n n a_n\ $ do not necessarily converge, by the counter-example above.
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This leads to the following question, for which I do not have an answer:
If $(a_n)$ is a decreasing real sequence and $\displaystyle\sum a_n $ converges, then does $\displaystyle\sum (-1)^n n a_n\ $ converge?
I was thinking we could apply Dirichlet's test or Cauchy's Condensation test somehow, but I don't quite see how.