Question

chứng minh rằng \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2002^2}+\frac{1}{2003^2}< 1\)

Answer 1

Đặt \(S=\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{2003^2}< \frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2002.2003}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2002}-\frac{1}{2003}\)

\(=1-\frac{1}{2003}< 1\)

Vậy S<1

Answer 2

bạn có thể giải rõ ra được ko

Answer 3

ta co 1/2^2<1/1*2+1/3^2+1/2*3+...+1/2003^2 1/2002*2003

1/2^2+1/3^2+...+1/2003^2<1/1*2+1/2*3+...+1/2002*2003

1/2^2+1/3^2+...+1/2003^2<1/1-1/2+1/2-1/3+...+1/2002-1/2003

1/2^2+1/3^2+...+1/2003^2<1-1/2003

1/2^2+1/3^2+...+1/2003^2<2002/2003<1

Vậy 1/2^2+1/3^2+...+1/2003^2<1