Question

[Phạm Vũ Thanh Nhàn]

So sánh tổng S=\(\dfrac{1}{5}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{41}\)+\(\dfrac{1}{42}\) và \(\dfrac{1}{2}\)

Answer 1

[Nguyễn Ngọc Minh Châu]

\(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\) và \(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)

Suy ra:

\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)