If $\sum\limits_{n=1}^\infty a_n$ converges and $a_n\geqslant 0$ for every $n$ then $\sum\limits_{n=1}^\infty a_n^{1/4}n^{-4/5}$ converges

Question

If $a_n \in \mathbb{R}^+$ and $\sum\limits_{n=1}^\infty a_n$ is convergent, $\sum\limits_{n=1}^\infty a_n^{1/4}n^{-4/5}$ converges.

How to prove this?

Answer 1

Hint Use Hölder's inequality for series.

Edit: since this answer was accepted, I think @cmi found the solution, so here is my version: Hölder's inequality gives

$$\sum _{n = 1}^{\infty } {a}_{n}^{\frac{1}{4}} \frac{1}{{n}^{\frac{4}{5}}} \leqslant {\left(\sum _{n = 1}^{\infty } {\left({a}_{n}^{\frac{1}{4}}\right)}^{p}\right)}^{\frac{1}{p}} {\left(\sum _{n = 1}^{\infty } {\left(\frac{1}{{n}^{\frac{4}{5}}}\right)}^{q}\right)}^{\frac{1}{q}}$$

when $\frac{1}{p}+\frac{1}{q} = 1$ and $p , q \geqslant 1$. We choose $p = 4$ and $q = \frac{4}{3}$. It gives

$$\sum _{n = 1}^{\infty } {a}_{n}^{\frac{1}{4}} \frac{1}{{n}^{\frac{4}{5}}} \leqslant {\left(\sum _{n = 1}^{\infty } {a}_{n}\right)}^{\frac{1}{4}} {\left(\sum _{n = 1}^{\infty } \frac{1}{{n}^{\frac{16}{15}}}\right)}^{\frac{3}{4}}$$

Both series on the RHS of this inequality converge, hence the series on the LHS also converges.

Answer 2

A different, more involved way without Hölder:

Let $N\geq 1$, $\alpha \in (1,\frac 65)$ and note that $$ \sum_{n=1}^N \frac{{a_n}^\frac{1}{4}}{n^\frac{4}{5}}\leq \sum_{n=1}^N \frac{{a_n}^\frac{1}{4}}{n^\frac{4}{5}}\mathbb{1}_{\large\left\{ \frac{1}{a_n^{1/2}n^{8/5}} < \frac{1}{n^{\alpha}} \right\}} + \sum_{n=1}^N \frac{{a_n}^\frac{1}{4}}{n^\frac{4}{5}}\mathbb{1}_{\large\left\{ \frac{1}{a_n^{1/2}n^{8/5}} \geq \frac{1}{n^{\alpha}} \right\}}$$

Let's deal with the first summand with Cauchy-Schwarz: $$\begin{align}\sum_{n=1}^N \frac{{a_n}^\frac{1}{4}}{n^\frac{4}{5}}\mathbb{1}_{\large \left\{ \frac{1}{a_n^{1/2}n^{8/5}} < \frac{1}{n^{\alpha}} \right\}} &\leq \left(\sum_{n=1}^Na_n \mathbb{1}_{\large \left\{ \frac{1}{a_n^{1/2}n^{8/5}} < \frac{1}{n^{\alpha}} \right\}} \sum_{n=1}^N \frac{1}{a_n^{1/2}n^{8/5}} \mathbb{1}_{\large \left\{ \frac{1}{a_n^{1/2}n^{8/5}} < \frac{1}{n^{\alpha}} \right\}} \right)^{1/2}\\ &\leq \left(\sum_{n=1}^N a_n \sum_{n=1}^N \frac{1}{n^{\alpha}} \mathbb{1}_{\large \left\{ \frac{1}{a_n^{1/2}n^{8/5}} < \frac{1}{n^{\alpha}} \right\}} \right)^{1/2} \\ &\leq \left(\sum_{n=1}^N a_n \sum_{n=1}^N \frac{1}{n^{\alpha}} \right)^{1/2} \\ &\leq \left(\sum_{n=1}^\infty a_n \sum_{n=1}^\infty \frac{1}{n^{\alpha}} \right)^{1/2} \end{align} $$

For the second summand, note that if $\displaystyle \frac{1}{a_n^{1/2}n^{8/5}} \geq \frac{1}{n^{\alpha}}$, then $\displaystyle \frac{{a_n}^\frac{1}{4}}{n^\frac{4}{5}}\leq \frac{1}{n^{8/5-\alpha/2}}$, so that

$$\sum_{n=1}^N \frac{{a_n}^\frac{1}{4}}{n^\frac{4}{5}}\mathbb{1}_{\large\left\{ \frac{1}{a_n^{1/2}n^{8/5}} \geq \frac{1}{n^{\alpha}} \right\}}\leq \sum_{n=1}^N \frac{1}{n^{8/5-\alpha/2}} \leq \sum_{n=1}^\infty \frac{1}{n^{8/5-\alpha/2}} $$

The last series converges because $\alpha$ was chosen so that $\alpha <\frac 65$.

Finally, $$\sum_{n=1}^N \frac{{a_n}^\frac{1}{4}}{n^\frac{4}{5}} \leq \left(\sum_{n=1}^\infty a_n \sum_{n=1}^\infty \frac{1}{n^{\alpha}} \right)^{1/2} + \sum_{n=1}^\infty \frac{1}{n^{8/5-\alpha/2}}$$ The right-hand side does not depend on $n$, we're done.

Answer 3

This isn't a complete answer, since it requires monotonically of $a_n$ If $a_n$ is monotonic, then

Define $$b_n=\frac{a_n^{1/4}}{n^{\frac{4}{5}}} , c_n=\frac{1}{n^{4/5+{1/4}}}$$

Then by the limit comparison test, define $ L :=\lim_{n\to \infty} \frac{b_n}{c_n}$, since $c_n$ converges, it is enough to show that $L \ne \infty $, (notice if $L=0$, $b_n$ still converges.)

$$\lim_{n\to \infty} \frac{\frac{a_n^{1/4}}{n^{\frac{4}{5}}}}{\frac{1}{n^{\frac{4}{5}+\frac{1}{4}}}}=\lim_{n\to \infty}a_n^{\frac{1}{4}}n^{\frac{1}{4}}=\lim_{n\to \infty} (a_nn)^{\frac{1}{4}}=0$$

Since $a_n$ converges, $(a_nn)^C \to 0$ for every $C$, (I will leave this part for you to prove, hint: Cauchy convergence test)