Clarence Hines

2022-01-24

For ${a}_{n}\ge 0$, and $\sum {a}_{n}$ convergent, show that $\sum \frac{\sqrt{{a}_{n}}}{{n}^{p}}$ is also convergent for $p>\frac{1}{2}$?

Dominique Green

Expert

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

Allison Compton

Expert

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

RizerMix

Expert

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 ${a}_{n}=\frac{1}{n\left(\mathrm{log}n{\right)}^{3/2}}$, then $\sum _{n}{a}_{n}$ is convergent and ${a}_{n}^{1/2}{n}^{-1/2}=\frac{1}{n\left(\mathrm{log}n{\right)}^{3/4}}$ and the series $\sum _{n}\frac{1}{n\left(\mathrm{log}n{\right)}^{3/4}}$ is divergent.