If $a_n$ is sequence of positive numbers such that $\sum a_n$ converges, then does $\sum \frac {{(a_n)}^{\frac 14}}{n^{\frac 45}}$ converge?

Hint:

For $u,v,w,t$ real and $\geq 0$, you have $$uvwt\leq \frac{u^4+v^4+w^4+t^4}{4}$$

Apply this with $u=a_n^{1/4}$, $\displaystyle v=w=t=\frac{1}{n^{4/15}}$

You can also use Holder's inequality: Take $p= 4, q=4/3,$ so that $1/p + 1/q=1.$ Holder then gives

$$\sum a_n^{1/4}n^{-4/5} \le \left [\sum (a_n^{1/4})^p)\right ]^{1/p}\cdot \left [\sum (n^{-4/5})^q \right ]^{1/q} = \left [\sum a_n \right ]^{1/4}\cdot \left [\sum n^{-16/15}\right ]^{3/4}.$$

On the right, the first factor is finite by assumption, and since $-16/15 < -1,$ the second factor is finite as well.