Đặt \(A=\dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+...+\dfrac{5}{49}\)
\(=5\left(\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{49}\right)\)
Đặt \(B=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{49}\Rightarrow\) Chứng minh \(\dfrac{3}{5}< B< \dfrac{8}{5}\)
Ta có:
\(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{29}< \dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\) \(=\dfrac{1}{2}\)
\(\dfrac{1}{30}+\dfrac{1}{31}+...+\dfrac{1}{34}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}\) \(=\dfrac{1}{2}\)
\(\dfrac{1}{35}+\dfrac{1}{36}+...+\dfrac{1}{49}< \dfrac{1}{35}+\dfrac{1}{35}+...+\dfrac{1}{35}\) \(=\dfrac{3}{7}< \dfrac{3}{5}\)
\(\Rightarrow B< \dfrac{1}{2}+\dfrac{1}{2}+\dfrac{3}{5}=\dfrac{8}{5}\)
\(\Leftrightarrow A< 8\left(1\right)\)
Lại có:
\(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{24}>\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{25}\) \(=\dfrac{1}{5}\)
\(\dfrac{1}{25}+\dfrac{1}{26}+...+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}\) \(=\dfrac{1}{5}\)
\(\dfrac{1}{30}+\dfrac{1}{31}+...+\dfrac{1}{37}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}\) \(=\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{37}>\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Rightarrow B>\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{37}>\dfrac{3}{5}\)
\(\Leftrightarrow A>3\left(2\right)\)
Kết hợp \(\left(1\right)\) và \(\left(2\right)\) suy ra:
\(3< \dfrac{5}{20}+\dfrac{5}{21}+\dfrac{5}{22}+...+\dfrac{5}{49}< 8\) (Đpcm)