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Calculus Level 4

n = 4 1000000 1 n 3 = ? \displaystyle \large \left \lfloor \sum _{n=4}^{1000000} \frac{1}{\sqrt[3]{n}} \right \rfloor = \ ? \


The answer is 14996.

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1 solution

Chew-Seong Cheong
Sep 24, 2018

We can estimate S = k = a b 1 n 3 \displaystyle S = \sum_{k=a}^b \frac 1{\sqrt[3] n} with I = a 1 b 1 1 x 3 d x \displaystyle I = \int_{a_1}^{b_1} \frac 1{\sqrt[3] x} dx as follows:

4 1000000 1 x 3 d x < k = a b 1 n 3 < 3 1000001 1 x 3 d x Note that 1 x 3 is a convex function. 3 x 2 3 2 4 1000000 < k = a b 1 n 3 < 3 x 2 3 2 3 1000001 14996.220 < k = a b 1 n 3 < 14996.890 \begin{aligned} \int_4^{1000000} \frac 1{\sqrt[3] x} dx < & \sum_{k=a}^b \frac 1{\sqrt[3] n} < \int_3^{1000001} \frac 1{\sqrt[3] x} dx & \small \color{#3D99F6} \text{Note that }\frac 1{ \sqrt[3]x} \text{ is a convex function.} \\ \frac {3x^\frac 23}2\bigg|_4^{1000000} < & \sum_{k=a}^b \frac 1{\sqrt[3] n} < \frac {3x^\frac 23}2\bigg|_3^{1000001} \\ 14996.220 < & \sum_{k=a}^b \frac 1{\sqrt[3] n} < 14996.890 \end{aligned}

Therefore, k = a b 1 n 3 = 14996 \displaystyle \left \lfloor \sum_{k=a}^b \frac 1{\sqrt[3] n} \right \rfloor = \boxed{14996} .

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