Question

Cho A=1/2^2+1/3^2+1/4^2+.....+1/100^2

CMR:A>1/2

Answer 1

ta có:1/2^2=1/4
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
...
1/100^2<1/99.100=1/99-1/100
=> A=1/2^2+1/3^2+1/4^2+.....+1/100^2<1/4+1/2-1/3+1/3-1/4+...+1/99-1/100
<1/4+1/2-1/100<1/2

Answer 2

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{100^2}\)

\(< \frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{99\cdot100}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}\)

\(< \frac{1}{2}-\frac{1}{100}\)

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