Question

Chứng mình rằng:

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

Answer 1

ta thấy:

\(\frac{1}{2^2}< \frac{1}{1.2}\)

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

\(\frac{1}{4^2}< \frac{1}{3.4}\)

...

\(\frac{1}{99^2}< \frac{1}{98.99}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{99^2}< \frac{1}{1}-\frac{1}{99}< 1\)

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