Đặt \(A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}\)
\(=\left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\)
Mặt khác:
\(\dfrac{7}{12}=\dfrac{20}{60}+\dfrac{20}{80}\)
mà \(\left\{{}\begin{matrix}\dfrac{20}{60}< \left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)\\\dfrac{20}{80}< \left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\end{matrix}\right.\)
⇒ \(\dfrac{7}{12}< A\) (1)
Ta có:
\(\dfrac{5}{6}=\dfrac{20}{40}+\dfrac{20}{60}\)
mà \(\left\{{}\begin{matrix}\dfrac{20}{40}>\left(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{60}\right)\\\dfrac{20}{60}>\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{80}\right)\end{matrix}\right.\)
⇒ \(A< \dfrac{5}{6}< 1\)(2)
Từ (1) và (2)
⇒ \(\dfrac{7}{12}< A< 1\) (đpcm)
