Question

Chứng minh rằng:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}<1\)

Answer 1

Ta có :

\(\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{4^2}=\frac{1}{4.4}<\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)

..............

\(\frac{1}{99^2}=\frac{1}{99.99}<\frac{1}{98.99}=\frac{1}{98}-\frac{1}{99}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}<\frac{98}{99}<1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}<1\left(đpcm\right)\)