Cube Rooted Till 2019

We can estimate the value of $\displaystyle \sum_{n=1}^{2019} \frac 1{\sqrt[3]n}$ using $\displaystyle \int_a^b \frac 1{\sqrt[3]x} dx$ with appropriate limits $a$ and $b$ . Since $f(x) = \dfrac 1{\sqrt[3]x}$ is convex, the area under the curve is large when the curve is above the bar than when it is under the bar (see figure). Therefore, we have:

$\begin{aligned} \frac 1{2\sqrt[3]1} + \int_1^{2019} \frac 1{\sqrt[3] x} dx+ \frac 1{2\sqrt[3]{2019}} < & \sum_{n=1}^{2019} \frac 1{\sqrt[3]n} < \int_{0.5}^{2019.5} \frac 1{\sqrt[3] x} dx \\ \frac 1{2\sqrt[3]1} + \frac 32 x^\frac 23 \bigg|_1^{2019} + \frac 1{2\sqrt[3]{2019}} < & \sum_{n=1}^{2019} \frac 1{\sqrt[3]n} < \frac 32 x^\frac 23 \bigg|_{0.5}^{2019.5} \\ 238.6553713 < & \sum_{n=1}^{2019} \frac 1{\sqrt[3]n} < 238.7104288 \end{aligned}$

$\implies \displaystyle \left \lfloor \sum_{n=1}^{2019} \frac 1{\sqrt[3]n} \right \rfloor = \boxed{238}$ .