\(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{16}{30}>\frac{15}{30}=\frac{1}{2}\)

S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)

suy ra S<1/5+1/12.3+1/60.3

S<1/5+1/4+1/20

S<1/2 

S<1/2

S=\(\frac{1}{5}\)+(\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\)) + (\(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\))

=> S< \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)

S<\(\frac{1}{5}+\frac{1}{4}+\frac{1}{20}\)

=> S< \(\frac{1}{2}\)

Vậy S<\(\frac{1}{2}\)

Ta có :

\(S=\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+.......+\frac{1}{62}+\frac{1}{63}+\frac{1}{64}\)

\(\Rightarrow S< \frac{1}{17}+\frac{1}{17}+......+\frac{1}{17}+\frac{1}{17}+\frac{1}{17}\)

\(\Rightarrow S< \frac{1}{17}.48\)

\(\Rightarrow S< \frac{48}{17}\)

\(\Rightarrow S< 2\)( 1 ) 

Lại có :

\(S>\frac{1}{64}+\frac{1}{64}+.........+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}\)

\(\Rightarrow S>\frac{1}{64}.48\)

\(\Rightarrow S>\frac{3}{4}\)( 2 ) 

Từ ( 1 ) và ( 2 ) suy ra : \(\frac{3}{4}< S< 2\)

Vậy \(1< S< 2\left(ĐPCM\right)\)

S lớn hơn

Ta có: 

\(\frac{1}{5}=\frac{1}{5}\)

\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}

Ta có: \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)