Ta có:\(P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
Mặt khác:
\(\dfrac{1}{2^2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)
.....
\(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
\(\Rightarrow P>\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{100.101}\)
\(\Rightarrow P>\dfrac{1}{2}-\dfrac{1}{101}\)
\(\Rightarrow P>\dfrac{99}{202}=\dfrac{1485}{3030}>\dfrac{1414}{3030}=\dfrac{7}{15}\)
\(\Rightarrow P>\dfrac{7}{15}\)
\(P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(P=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)
\(P>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{100.101}\)
\(P>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(P>\dfrac{1}{2}-\dfrac{1}{101}\)
\(P>\dfrac{101}{202}-\dfrac{2}{202}\)
\(P>\dfrac{99}{202}\)
Vì \(\dfrac{99}{202}>\dfrac{7}{15}\)
\(\Rightarrow P>\dfrac{99}{202}>\dfrac{7}{15}\)
\(P>\dfrac{7}{15}\)
