Question

Chung minh

\(A=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+.....+\dfrac{1}{20}\)\(< \dfrac{1}{2}\)

Answer 1

Ta có :

\(A=\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{1}{12}+\dfrac{1}{12}+...+\dfrac{1}{12}\left(6PS\right)\)

Mà\(\dfrac{1}{12}+\dfrac{1}{12}+...+\dfrac{1}{12}=6.\dfrac{1}{12}=\dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{1}{2}\)

Answer 2

\(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{1}{2}\\ \dfrac{1}{10}>\dfrac{1}{12}\\ \dfrac{1}{12}=\dfrac{1}{12}\\ ...\\ \dfrac{1}{20}< \dfrac{1}{12}\)

⇒Cộng 2 vế, ta có:

\(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+...+\dfrac{1}{20}< \dfrac{6}{12}=\dfrac{1}{2}\)

Vậy A<\(\dfrac{1}{2}\)