Convergence of $\sum \frac{a_n }{\sqrt{n}}$ given that $\sum a_n^2$ converges

Hypothesis: If $\sum_{n=1}^{\infty} a_n^2$ converges then $\sum_{n=1}^{\infty} \frac{a_n}{\sqrt{n}}$ also converges.

I'm looking for a counterexample (or proof) to this claim. The only assumption is $a_n \in \mathbb{R}$. Here it was proved that convergence of $\sum a_n^2$ implies convergence of $\sum \frac{a_n}{n^p}$ for $p=1$. I noticed it can be easily generalised for $p>\frac{1}{2}$ since: $$ 0 \leq (a_n - \frac{1}{n^p})^2 = a_n^2 - 2\frac{a_n}{n^p} + \frac{1}{n^{2p}} \implies \frac{a_n}{n^p} \leq \frac{1}{2}(a_n^2 + \frac{1}{n^{2p}}) $$ On the other hand, for $p<\frac{1}{2}$ there is a counterexample: $a_n = \frac{1}{n^{1-p}}$. But it doesn't apply to $p= \frac{1}{2}$, since then $\sum a_n^2 = \sum \frac{1}{n}$, which diverges. So no p-series can be a counter and also Cauchy-Schwarz doesn't help.

Solution 1:

Let $a_n=\frac{1}{\sqrt{n}\log n}$. Then $\sum_{n=2}^\infty a_n^2 = \sum_{n=2}^\infty \frac{1}{n\log^2 n}$ which converges by Cauchy's condensation test, but $$\sum_{n=2}^\infty\frac{a_n}{\sqrt{n}} = \sum_{n=2}^\infty \frac{1}{n\log n}$$ and this diverges, again by Cauchy's condensation test.