Floored Cube Roots

Calculus Level 4

$\large \left \lfloor \dfrac{1}{\sqrt[3]{1}} + \dfrac{1}{\sqrt[3]{2}} + \dfrac{1}{\sqrt[3]{3}} + \cdots + \dfrac{1}{\sqrt[3]{2019}} \right \rfloor = \, ?$

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 238.

1 solution

Manuel Kahayon
Aug 12, 2016

Let $S = \frac{1}{\sqrt[3]{1}} + \frac{1}{\sqrt[3]{2}} + \frac{1}{\sqrt[3]{3}} + \cdots + \frac{1}{\sqrt[3]{2019}}$ . We then know that

$\large S \approx \int _1^{2020} x^{-\frac{1}{3}} dx = \frac{3}{2} (2020^\frac{2}{3} - 1^\frac{2}{3}) = 238.19$

Therefore, $S \approx 238.19$ . Therefore, $\lfloor S \rfloor = \lfloor 238.19 \rfloor = \boxed{238}$ .

Floored Cube Roots

The answer is 238.

1 solution

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