How to prove convergence of $\sum\limits_{n=1}^\infty \frac{\sqrt a_n}{n^\alpha}$

Question

Assume that $a_n >0$, $n\in\mathbb{N}$, and that $\sum\limits_{n=1}^\infty a_n $ converges. Show that for $\alpha>1/2$ the series $$\sum\limits_{n=1}^\infty \frac{\sqrt a_n}{n^\alpha}$$ converges as well

I think we can use Abel's convergence test but I'm not sure if that is the way to go

Answer 1

By Cauchy–Schwarz inequality we have

$$\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n^{\alpha}}\leq \sqrt{\sum_{n=1}^\infty a_n}\,\sqrt{\sum_{n=1}^\infty \frac{1}{n^{2\alpha}}}$$

and

$$\sum_{n=1}^\infty \frac{1}{n^{2\alpha}}$$

converges if and only if $2\alpha >1 \implies \alpha >\frac12$.

Answer 2

hint: The Cauchy-Schwarz inequality gives $$\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n}\leq \sqrt{\sum_{n=1}^\infty a_n}\,\sqrt{\sum_{n=1}^\infty \frac{1}{n^2}}<\infty.$$

Answer 3

Just to give an elementary proof (not relying on Cauchy-Schwartz), let $\alpha={1+p\over2}$. Then

$$\begin{align} \left|\sqrt{a_n/n^{1+p}}-1/n^{1+p}\right| &={\left|a_n/n^{1+p}-1/n^{2+2p}\right|\over\sqrt{a_n/n^{1+p}}+1/n^{1+p}}\\ &\le{\left|a_n/n^{1+p}-1/n^{2+2p}\right|\over1/n^{1+p}}\\ &=\left|a_n-1/n^{1+p}\right|\\ &\le a_n+1/n^{1+p} \end{align}$$

and thus

$$\begin{align} \left|\sqrt{a_n/n^{1+p}}\right| &=\left|\sqrt{a_n/n^{1+p}}-1/n^{1+p}+1/n^{1+p}\right|\\ &\le\left|\sqrt{a_n/n^{1+p}}-1/n^{1+p}\right|+1/n^{1+p}\\ &\le a_n+2/n^{1+p} \end{align}$$

and so

$$\sum_{n=1}^\infty{\sqrt{a_n}\over n^\alpha}\le\sum_{n=1}^\infty a_n+2\sum_{n=1}^\infty{1\over n^{1+p}}\lt\infty$$