P/s:Bạn chép sai đề rồi nhea để mình sủa lại đề cho đúng nha!!!
Chứng minh rằng :
\(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
= \(\dfrac{1}{2^2}+\dfrac{1}{3^2} +...+\dfrac{1}{100}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{100}\)
=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{99}{100}< 1\)
=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1\)
