Question

cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2003^2}\)

CTR \(\frac{1}{3}< A< 1\)

Answer 1

Ta có \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};...;\frac{1}{2003^2}< \frac{1}{2002\cdot2003}\)

Suy ra \(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2002\cdot2003}\)

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

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

\(\Rightarrow A< 1\)

Answer 2

Ta có \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2003^2}< \frac{1}{2002.2003}\)

Suy ra \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2002.2003}\)

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

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

\(\Rightarrow A< 1\)

Answer 3

ctr \(\frac{1}{3}< A< 1\) mà bạn ơi

Answer 4

Ta có  \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2003^2}>\frac{1}{4}+\frac{1}{9}\)

Mà \(\frac{1}{4}+\frac{1}{9}=\frac{13}{36}>\frac{12}{36}=\frac{1}{3}\)

\(\Rightarrow A>\frac{1}{3}\)

Vậy \(\frac{1}{3}< A< 1\)