อ้างอิง:
ข้อความเดิมเขียนโดยคุณ -InnoXenT-
..4. Find $\left\lfloor\ \frac{A}{4} \right\rfloor$ , if $A = \sum_{n = 1000}^{1000000} \frac{1}{\sqrt[3]{n}} $
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ไม่แน่ใจว่าทำแบบนี้ได้รึเปล่า
from
$\frac{1}{\sqrt[3]{1000}}+\frac{1}{\sqrt[3]{1000}}+...+\frac{1}{\sqrt[3]{1000}}> \sum_{n = 1000}^{1000000} \frac{1}{\sqrt[3]{n}}>frac{1}{\sqrt[3]{1000000}}+\frac{1}{\sqrt[3]{1000000}}+...+\frac{1}{\sqrt[3]{1000000}}$
$\frac{999001}{10}>\sum_{n = 1000}^{1000000} \frac{1}{\sqrt[3]{n}}>\frac{999001}{100}$
$99900.1>\sum_{n = 1000}^{1000000} \frac{1}{\sqrt[3]{n}}>9990.1$
$99900.1>A>9990.1$
$24975...>\frac{A}{4}>2497...$