Convergence of $\sum \frac{\sqrt{a_n}}{n^p}$

1). Lemma $a_n,b_n \geq 0$, $\sum a_n, \sum b_n$ convergent, then $$\sum \sqrt{a_nb_n}$$ is convergent, too. This is because

$$\sqrt{a_nb_n}\leq\frac12 (a_n+b_n)$$

Now $b_n=\frac1{n^{2p}}$, $\sum b_n$ is convergent

2). a counterexample $$a_n=\frac 1{n(\log n)^2}$$

we get

$$\sqrt{\frac{a_n}{n}}=\frac 1{n\log n} $$

so $\sum \sqrt{\frac{a_n}{n}}$ is divergent.

Since $\frac{\sqrt{a_n}}{n^p}$ is a sequence with non-negative terms, then the series $$ \sum_{n=1}^\infty \frac{\sqrt{a_n}}{n^p}, $$ converges if and only if it is bounded.

Cauchy-Schwarz provides that $$ \left( \sum_{n=1}^N \frac{\sqrt{a_n}}{n^p} \right)^{\!2}\le \left(\sum_{n=1}^N a_n\right)\left(\sum_{n=1}^N\frac{1}{n^{2p}}\right)\le \left(\sum_{n=1}^\infty a_n\right)\left(\sum_{n=1}^\infty\frac{1}{n^{2p}}\right), $$ and as the right-hand side is bounded for $2p>1$, then so is the sequence of the partial sums $\displaystyle\sum_{n=1}^N \frac{\sqrt{a_n}}{n^p}$.

Therefore, the series $\displaystyle\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n^p}$ converges for $p>1/2$.

If $p>1/2$, convergence of $\sum_n a_n^{1/2}n^{-p}$ follows from Cauchy-Schwarz inequality and the fact that $\sum_n n^{-r}$ is convergence for $r\gt 1$.

If we take $a_n=\frac 1{n(\log n)^{3/2}}$, then $\sum_n a_n$ is convergent and $$a_n^{1/2}n^{-1/2}=\frac 1{n(\log n)^{3/4}}$$ and the series $\sum_n\frac 1{n(\log n)^{3/4}}$ is divergent.