Question

[Nguyễn Thái Bảo]

tính tổng 1/2²+1/3²+1/4²+...+1/100²<1

Answer 1

[Nguyễn Quang Tâm]

Gọi tổng \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\) là A.

Theo bài ra, ta có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\\ A=\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot3}+\dfrac{1}{4\cdot4}+...+\dfrac{1}{100\cdot100}\\ \Rightarrow A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ \Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{1}-\dfrac{1}{100}\\ \Rightarrow A< \dfrac{99}{100}< 1\)

Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)