Question

Chứng minh 1/2^2 + 1/3^2 +...+ 1/99^2 <  3/4

Answer 1

      \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}\)
    = \(\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{99.99}\)
A < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)
A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
A < \(1-\dfrac{1}{100}\)
A < \(\dfrac{100}{100}-\dfrac{1}{100}\)
A < \(\dfrac{99}{100}\)<\(\dfrac{3}{4}\)
Do đó \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}< \dfrac{3}{4}\)