I am trying to determine to whether the series
$\sum _{n=0}^{\infty }\:\left(-1\right)^n\frac{1}{\sqrt{n}\left(ln\left(n\right)^{2021}\right)}$
is conditionally convergent, absolutely convergent, or divergent. So far using p-series testing I believe that $\frac{1}{\sqrt{n}\left(ln\left(n\right)^{2021}\right)}$ is decreasing for p=1/2 I was wondering if we can possibly argue that this series is conditionally divergent or not?
You should take $\sum_{n=2}^\infty \cdots$. $\sqrt{n}\ln(n)^{2021}=2021\sqrt{n}\ln(n)$ is an increasing function of $n$ which goes to $+\infty$ as $n$ increases (can be verified easily), so $\frac{1}{2021\sqrt{n}\ln(n)}$ is decreasing and goes to $0$ as $n$ increases. Hence it's conditionally convergent.
$ \sqrt{n} <n$ so $\frac{1}{2021\sqrt{n}\ln(n)}>\frac{1}{2021 ~n\ln(n)}$.
$\frac{1}{2021\sqrt{n}\ln(n)}$ diverges since $\frac{1}{2021 ~n\ln(n)}$ diverges (which is known). Given series is not absolutely convergent.