Question

Chứng tỏ rằng :

\(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2} \) + ... + \(\dfrac{1}{100^2}\) < 1

Answer 1

Gọi biểu thức trên là A:

Có 1/2²=1/2.2<1/1.2; 1/3²=1/3.3<1/2.3;.....;1/100²=1/100.100<1/99.100

=>A<1/1.2+1/2.3+1/3.4+....+1/99.100

=>A< 1/1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100

=>A<1-1/100=99/100

=>A<1 vì 99/100<1 mà A<99/100

Answer 2

\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)

\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{1}\)-\(\dfrac{1}{100}\)=\(\dfrac{100}{100}\)-\(\dfrac{1}{100}\)=\(\dfrac{99}{100}\)

Vì \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{99}{100}\)<1 nên \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<1

Vậy \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<1.