Ta có: \(\frac{1}{51}<\frac{1}{50}\)

\(\frac{1}{52}<\frac{1}{50}\)

...

\(\frac{1}{91}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{91}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{42}{50}\)

=>\(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{91}<\frac{42}{50}+\frac{1}{50}\)

=>\(K<\frac{43}{50}<1\) (1)

Ta có: \(\frac{1}{50}>\frac{1}{70}\)

\(\frac{1}{51}>\frac{1}{70}\)

...

\(\frac{1}{69}>\frac{1}{70}\)

Do đó: \(\frac{1}{50}+\frac{1}{51}+\cdots+\frac{1}{69}>\frac{1}{70}+\frac{1}{70}+\cdots+\frac{1}{70}=\frac{20}{70}\)

=>\(\frac{1}{50}+\frac{1}{51}+\cdots+\frac{1}{70}>\frac{20}{70}+\frac{1}{70}=\frac{21}{70}=\frac{3}{10}\) (2)

Ta có: \(\frac{1}{71}>\frac{1}{91};\frac{1}{72}>\frac{1}{91};\ldots;\frac{1}{90}>\frac{1}{91};\frac{1}{91}=\frac{1}{91}\)

Do đó: \(\frac{1}{71}+\frac{1}{72}+\cdots+\frac{1}{91}>\frac{1}{91}+\frac{1}{91}+\cdots+\frac{1}{91}=\frac{21}{91}=\frac{3}{13}\) (3)

Từ (2),(3) suy ra \(K>\frac{3}{10}+\frac{3}{13}=3\cdot\left(\frac{1}{10}+\frac{1}{13}\right)=3\cdot\frac{23}{130}=\frac{69}{130}\)

=>\(K>\frac{65}{130}=\frac12\) (4)

Từ (1),(4) suy ra 1/2<K<1