Convergence of $\sum(-1)^k\frac{(\ln k)^p}{k^q}$ where $p,q$ in positive $\mathbb{R}$

Question

For any $p, q$ in positive $\mathbb{R}$ $$\sum_{k=2}^{\infty}(-1)^k\frac{(\ln k)^p}{k^q}$$

I want to Use alternative series test for convergence but I'm struggling to verify that $\frac{(\ln k)^p}{k^q}$ is a monotone decreasing. roughly, since $k$ is faster then $\ln k$, it might be decreasing but it's just a guess

Answer 1

• If $q>1$ then let $q'$ such that $q>q'>1$ and since

$$(\ln k)^p=_\infty o(k^{q'})$$ then we see that the given series is absolutely convergent by comparison with a Riemann series.

• If $0<q\le1$ and using the function

$$f(x)=\frac{(\ln x)^p}{x^q}$$ so $$f'(x)=\frac{p x^{q-1}(\ln x)^{p-1}-qx^{q-1}(\ln x)^p}{(x^q)^2}=\frac{ x^{q-1}(\ln x)^{p-1}(p-q\ln x)}{(x^q)^2}$$ we see that the derivative is negative for large $x$ and then $f$ is decreasing and the sequence converges to $0$. We apply the Leibniz criterion.

Remark $\quad$ For $0<q\le1$ the series isn't absolutely convergent since $$\frac{(\ln k)^p}{k^q}\ge\frac1{k^q}$$