Ta có
\(P< \frac{1}{4.5}+\frac{1}{5.6}+-..+\frac{1}{99.100}\)
\(\Rightarrow P< \frac{1}{4}-\frac{1}{5}+-.+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow P< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(\Rightarrow P< \frac{1}{4}\left(1\right)\)
\(p>\frac{1}{5^2}+\frac{1}{6.7}+-+\frac{1}{100.101}\)
\(P>\frac{1}{5^2}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(P>\frac{1}{6}+\frac{1}{25}-\frac{1}{101}\)
Ta thấy
\(\frac{1}{25}>\frac{1}{101}\Rightarrow\frac{1}{25}-\frac{1}{101}>0\)
Đặt \(M=\frac{1}{25}-\frac{1}{101}\)
\(\Rightarrow P>\frac{1}{6}+M>\frac{1}{6}\)
\(\Rightarrow P>\frac{1}{6}\left(2\right)\)
Tự (1) và (2)
\(\Rightarrow\frac{1}{6}< p< \frac{1}{4}\)
