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Due to Euler- Machaurin summation we have $\sum_{k=1}^{2020}\frac{1}{\sqrt[4]{k}} =\zeta\left(\frac{1}{4}\right)+\frac{4\sqrt[4]{2020^3}}{3}+\frac{1}{4}\int_{2020}^{\infty}\frac{\left\{x\right\}}{x^{5/4}}dx$ since $0 < \int_{2020}^{\infty}\frac{\left\{x\right\}}{x^{5/4}}dx < \int_{2020}^{\infty}\frac{dx}{x^{5/4}}=\frac{4\sqrt{2}}{\sqrt[4]{505}}$ so the approximated sum $\sum_{k=1}^{2020}\frac{1}{\sqrt[4]{k}} \approx \zeta\left(\frac{1}{4}\right)+\frac{4}{3}\sqrt[4]{2020^3}+\frac{\sqrt{2}}{\sqrt[4]{2020}}\sim 401.35526$ so floor value we have $401$ .