For brevity, let
[tex]S_i = \displaystyle \sum_{k=1}^n \frac1{\sqrt[i]{k}}[/tex]
Rewrite the limit as
[tex]\displaystyle \lim_{n\to\infty} \frac{{S_5}^{\frac23}}{{S_3}^{\frac45}} = \lim_{n\to\infty} \frac{n^{\frac23} \left(\frac{S_5}n\right)^{\frac23}}{n^{\frac45} \left(\frac{S_3}n\right)^{\frac45}} = \lim_{n\to\infty} \frac1{n^{\frac2{15}}} \times \frac{\left(\lim\limits_{n\to\infty}\left(\frac1n S_5\right)\right)^{\frac23}}{\left(\lim\limits_{n\to\infty}\left(\frac1n S_3\right)\right)^{\frac45}}[/tex]
The two limits involving S are equivalent to definite integrals,
[tex]\displaystyle \lim_{n\to\infty} \frac1n S_i = \lim_{n\to\infty} \frac1n \sum_{k=1}^n \frac1{\sqrt[i]{\frac kn}} = \int_0^1 \frac1{\sqrt[i]{x}} \, dx = \frac i{i-1}[/tex]
Then the end result is
[tex]\displaystyle 0 \times \frac{\left(\frac54\right)^{\frac23}}{\left(\frac32\right)^{\frac45}} = \boxed{0}[/tex]
