Convergence of $\sum \frac{\sqrt{a_n}}{n^p}$ For $a_n \geq 0$, and $\sum a_n$ convergent, show that  $\sum \frac{\sqrt{a_n}}{n^p}$ is also convergent for $p > 1/2$?
What bugs me more is why isn't $\sum \sqrt{\frac{a_n}{n}}$ convergent??
Clearly it converges for $p > 1$, and $a_n$ somehow helps out in $p \in (1/2,1)$.
Only hints are welcomed
Thanks in advance..
 A: 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.
A: 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 $p>1/2$.
A: 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. 
