Clarence Hines

Answered

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 $p>\frac{1}{2}$ ?

Answer & Explanation

Dominique Green

Expert

2022-01-25Added 11 answers

1). Lemma $a}_{n},\text{}{b}_{n}\ge 0,\text{}\sum {a}_{n},\text{}\sum {b}_{n$ convergent, then

$\sum \sqrt{{a}_{n}{b}_{n}}$

is convergent, too. This is because

$\sqrt{{a}_{n}{b}_{n}}\le \frac{1}{2}({a}_{n}+{b}_{n})$

Now$b}_{n}=\frac{1}{{n}^{2p}},\text{}\sum {b}_{n$ 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.

is convergent, too. This is because

Now

2). a counterexample

we get

so

Allison Compton

Expert

2022-01-26Added 16 answers

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 $p>\frac{1}{2}$ .

converges if and only if it is bounded.

Cauchy-Schwarz provides that

and as the right-hand side is bounded for

Therefore, the series

RizerMix

Expert

2022-01-27Added 437 answers

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

Most Popular Questions