Find all values of $p$ such that the series $\displaystyle\sum_{j=1}^{\infty}\dfrac{(-1)^j\cdot\log(j)}{j^p}$ is conditionally convergent.
I know that the alternating series test states that if a series has the form $\sum_{j=1}^{\infty}(-1)^ja_j$ where $a_j$ is a positive sequence that decreases monotonically to zero, then the series is convergent if and only if $a_{j+1} \leq a_j$ for all $j$ and $\lim_{j \to \infty}a_j = 0$.
So from that, I think that in this case, we have $a_j = \frac{\log(j)}{j^p}$. To satisfy the conditions of the alternating series test, we need to ensure that $a_{j+1} \leq a_j$ for all $j$ and $\lim_{j \to \infty}a_j = 0$. But I'm not sure if this is correct, and I anyways am unsure of how I can continue the problem from there. Can someone help me?