Lê Quốc Trần Anh Coordinator
12/07/2017 at 08:57-
We have :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\) (1)
In another case , we have :
\(\dfrac{1}{2^2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)
................
\(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{100^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{100.101}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{99}{202}\) (2)
From (1) and (2)
=> \(\dfrac{99}{100}>\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{100^2}>\dfrac{99}{202}\)
Lê Quốc Trần Anh selected this answer.