Ta có :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{200^2}\)
\(\Rightarrow\) \(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{199.200}\)
\(\Rightarrow\) \(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}=1-\dfrac{1}{200}\)
\(\Rightarrow\) \(A< 1\) (đpcm)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2},\dfrac{1}{3^2}< \dfrac{1}{2.3},...,\dfrac{1}{200^2}< \dfrac{1}{199.200}\)
⇒A<\(\dfrac{1}{1.2}.\dfrac{1}{2.3}.\dfrac{1}{3.4}.....\dfrac{1}{199.200}\)
A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
A<\(1-\dfrac{1}{200}\)
A\(< \)\(\dfrac{199}{200}\)\(< 1\)(đpcm)
a) Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{200^2}< \dfrac{1}{199\cdot200}\)
Do đó: \(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{199\cdot200}\)
\(\Leftrightarrow A< 1-\dfrac{1}{200}< 1\)(đpcm)
Ta có: 11.2.12.3.13.4.....1199.20011.2.12.3.13.4.....1199.200
A<1−12001−1200
A<<