Different tests of convergence

Question

For a non-negative sequence $(x_n)_{n\in\mathbb N}$, consider the following two criteria: $$\sum_{n=1}^\infty \frac{(\log n)^2}{x_n^2} < \infty \tag{1}\label{1}$$ and $$\exists r>1 \colon \sum_{n=1}^\infty \frac{n^r}{x_n^{2r}} < \infty \tag{2}\label{2}.$$ I want to know which criterion is stronger. Up to now, I already found that \ref{1} does not imply \ref{2} by considering $x_n = \sqrt{n}(\log n)^2$. Is the other implication true or are there again counterexamles so that both criteria are indeed different?

Answer 1

I think first that your example show that (1) does not imply (2).

To prove that (2) imply (1), use the holder inequality with $q=r$, $p=r/(r-1)$, $\displaystyle x_k=\frac{(\log k)^2}{k}$, $\displaystyle y_k= \frac{k}{x_k^2}$ : One gets $$\sum_{k=1}^n \frac{(\log k)^2}{x_k^2}\leq (\sum_{k=1}^n \frac{(\log k)^{2p}}{k^{p}})^{1/p}(\sum_{k=1}^n \frac{k^q}{x_k^{2q}})^{1/q}$$ and as the two series $\displaystyle\frac{(\log k)^{2p}}{k^{p}}$ and $\displaystyle\frac{k^q}{x_k^{2q}}$ are convergent, we are done.