Đặt :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......................+\dfrac{1}{100^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
...........................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....................+\dfrac{1}{99.100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{100}< 1\)
\(\Rightarrow A< 1\rightarrowđpcm\)
Đặt \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}\)
.............................
\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(\Rightarrow S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow S< 1-\dfrac{1}{100}\)
\(\Rightarrow S< \dfrac{100}{100}-\dfrac{1}{100}\)
\(\Rightarrow S< \dfrac{99}{100}\)
Vì \(\dfrac{99}{100}< 1\)
\(\Rightarrow S< 1\)
Vậy \(S< 1\).
Đặt:
A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
Ta thấy :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\)
\(\dfrac{1}{100.100}< \dfrac{1}{99.100}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
Nhận xét :
\(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};\)
\(...;\dfrac{1}{99.100}=\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\)
\(\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{100}< 1\)
\(\Rightarrow A< 1\)
Ta có: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
=> \(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< 1\)
Ta thấy: \(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}< 1\)
\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}< 1\)
\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< \dfrac{1}{1}-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}< 1\)
\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< \dfrac{1}{1}-0-0-0-...-0-\dfrac{1}{100}< 1\)
\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< \dfrac{1}{1}-\dfrac{1}{100}< 1\)
\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< \dfrac{99}{100}< 1\)
=> \(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< 1\)
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
