Question

[nguyễn gia khánh]

B=1/2^2+1/3^2+1/4^2+...+1/100^2 CMR B<3/4

Answer 1

[Dũng Lê Trí]

Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(\frac{1}{1\cdot2}< \frac{1}{2^2}\)

\(\frac{1}{2\cdot3}< \frac{1}{3^2}\)

\(\frac{1}{3\cdot4}< \frac{1}{4^2}\)

...

\(\frac{1}{99\cdot100}< \frac{1}{100^2}\)

\(\Rightarrow B< A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(B< A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< A=1-\frac{1}{100}=\frac{99}{100}\)

\(\Rightarrow A=\frac{99}{100}>\frac{3}{4}\)

\(\Leftrightarrow B< \frac{3}{4}< A\)