Question

P=1/5²+1/6²+...+1/100²

 ^{chung minh rang 1/6<P<1/4}

Answer 1

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)

\(\Rightarrow\frac{1}{6}< p< \frac{1}{4}\)