A = (1/5)+(1/15)+(1/25)+...+(1/1985)=
1/5+1/3*5+1/5*5+1/7*5+.........+1/397*5
5A=1+1/3+1/5+1/7+.......+1/397
5A-1=1/3+1/5+1/7+.......+1/397
Đặt B= 1/3+1/5+1/7+.......+1/397
=>.......................
Tính đc B=5,06241 (lấy số gần bằng) => A= 1,2124 (lấy số gần bằng)
=> A < 9/20

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

lại chép câu trả lời của người khác rồi

Đặt \(A=\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+...+\dfrac{1}{1985}\)

\(A=\dfrac{1}{5}.\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{397}\right)\)

\(A=\dfrac{1}{5}.\left(1+\dfrac{1}{1+2}+\dfrac{1}{2+3}+...+\dfrac{1}{198+199}\right)\)

\(A=\dfrac{1}{5}.\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{198}-\dfrac{1}{199}\right)\)

\(A=\dfrac{1}{5}.\left(2-\dfrac{1}{199}\right)\)

\(A=\dfrac{397}{995}< \dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+...+\dfrac{1}{1985}< \dfrac{9}{20}\left(đpcm\right)\)