Question

chứng tỏ:

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{25^2}< 1\)

Answer 1

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{25^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{24.25}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=1-\dfrac{1}{25}< 1\)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{25^2}< 1\left(đpcm\right)\)

Vậy...