Question

Chứng minh:\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}>1\)

Answer 1

Ta có :

\(\frac{1}{10}>\frac{1}{20}\)

\(\frac{1}{11}>\frac{1}{20}\)

\(\frac{1}{12}>\frac{1}{20}\)     \(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+.....+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+....+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)(1)

.....

\(\frac{1}{19}>\frac{1}{20}\)

Ta có :

\(\frac{1}{20}>\frac{1}{30}\)

\(\frac{1}{21}>\frac{1}{30}\)

\(\frac{1}{22}>\frac{1}{30}\)      \(\Rightarrow\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+....+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)(2) 

........

\(\frac{1}{29}>\frac{1}{30}\)

Ta có :

\(\frac{1}{30}>\frac{1}{40}\)

\(\frac{1}{31}>\frac{1}{40}\)                \(\Rightarrow\frac{1}{30}+\frac{1}{31}+....+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)(3)

.........

\(\frac{1}{39}>\frac{1}{40}\)

Từ 1 , 2 , 3 ,

=> \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+.....+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)

=> ....... > 1 

Answer 2

1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2 
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3 
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4 
\(\Rightarrow\)1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1