For an≥0, and ∑an convergent, show that ∑annp is also convergent for p>12?

1). Lemma ${\displaystyle a_n,b_n\ge 0,\sum a_n,\sum b_n}$ convergent, then
${\displaystyle \sum \sqrt{a_n{b}_n}}$
is convergent, too. This is because
${\displaystyle \sqrt{a_n{b}_n}\le \frac{1}{2}(a_n+b_n)}$
Now ${\displaystyle b_n=\frac{1}{n^{2p}},\sum b_n}$ is convergent
2). a counterexample
${\displaystyle a_n=\frac{1}{n{\left(\log n\right)}^2}}$
we get
${\displaystyle \sqrt{\frac{a_n}{n}}=\frac{1}{n\log n}}$
so ${\displaystyle \sum \sqrt{\frac{a_n}{n}}}$ is divergent.

Since ${\displaystyle \frac{\sqrt{a_n}}{n^p}}$ is a sequence with non-negative terms, then the series
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{\sqrt{a_n}}{n^p}}$
converges if and only if it is bounded.
Cauchy-Schwarz provides that
${\displaystyle {\left(\sum _{n=1}^N\frac{\sqrt{a_n}}{n^p}\right)}^2\le \left(\sum _{n=1}^N{a}_n\right)\left(\sum _{n=1}^N\frac{1}{n^{2p}}\right)\le \left(\sum _{n=1}^{\mathrm{\infty }}{a}_n\right)\left(\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{2p}}\right)}$
and as the right-hand side is bounded for ${\displaystyle 2p>1}$, then so is the sequence of the partial sums ${\displaystyle \sum _{n=1}^N\frac{\sqrt{a_n}}{n^p}}$.
Therefore, the series ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{\sqrt{a_n}}{n^p}}$ converges for ${\displaystyle p>\frac{1}{2}}$.

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 $r>1$ 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.