S=1/2+1/3+1/4+1/5+1/6+1/7+1/8=

                                                     =481/280=1,717(857142)

                                                   => S<2

S=1+2+2^2+2^3+...+2^20

2.S=2+2^2+2^3+...+2^20+2^21

2.S-S=S=(2+2^2+2^3+....+2^21)-(1+2+2^2+...+2^20)

S=2^21-1

bây giờ so sánh 2^21-1 với 5.2^19

mà 2^21-1=2^19.2^2-1 hay 2^19 .4 -1 <2^19.5

=>S<2^19.5

\(S=1+2+2^2+2^3+\cdots+2^{20}\)

=> \(2S=2+2^2+2^3+2^4+\cdots+2^{21}\)

=> \(2S-S=\left(2+2^2+2^3+2^4+\cdots+2^{21}\right)-\left(1+2+2^2+2^3+\ldots+2^{20}\right)\)

=>\(S=2^{21}-1\)

Mà \(2^{21}-1=2^{19}.2^2-1\) hay \(2^{19}.4-1<2^{19}.5\)

=>\(S<2^{19}.5\)

Ta có:\(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\frac{1}{13}>\frac{1}{20};....;\frac{1}{19}>\frac{1}{20}\)

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)(Có 10 phân số \(\frac{1}{20}\))

\(\Rightarrow\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{10}{20}\)\(\Leftrightarrow S>\frac{10}{20}\)

Mà \(\frac{10}{20}=\frac{1}{2}\)nên

\(\Rightarrow S>\frac{1}{2}\)

A=(1/2+1/12+1/13+1/14+1/15)+(1/16+1/17+1/18+1/19+1/20)

Thay các phân số trong mỗi nhóm bởi phân số nhỏ nhất, ta có:

A> 1/15.5+1/20.5=1/3+1/4= 7/12>1/2

Suy ra A>1/2

Vậy A> 1/2

Sửa đề: \(S=\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{100}\)

Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{75}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{25}{50}=\frac12\) (1)

Ta có: \(\frac{1}{76}<\frac{1}{75};\frac{1}{77}<\frac{1}{75};\ldots;\frac{1}{100}<\frac{1}{75}\)

Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}<\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (2)

Từ (1),(2) suy ra \(\left(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}\right)+\left(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}\right)<\frac12+\frac13\)

=>\(S<\frac56\)

Vì 1/2<2/3;3/4<4/5;5/6<6/7;....;9999/10000<10000/10001

-->S<2/3.4/5.5/6.....10000/10001

Gọi 2/3.4/5.5/6.....10000/10001 là D và D>S

Có D=2/3.4/5.5/6.....10000/10001

-->S.D=(1/2 . 3/4 . 5/6.....9999/10000).(2/3.4/5.5/6.....10000/10001)

S.D=1/2.2/3.3/4.4/5.5/6.6.7.....9999/10000.10000/10001

S.D=1/10001

Vì S<D nên S.S<S.D hay S.S<1/10001

-->S<1/10001

mà 1/10001<1/100=0.01

-->S<1/100=0.01

-->S<0.01

Vậy

Có \(\frac{1}{2}< \frac{2}{3};\frac{2}{3}< \frac{3}{4};.....;\frac{9999}{10000}< \frac{10000}{10001}\)

\(\Rightarrow S^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{10000}{10001}\)

\(\Rightarrow S^2< \frac{1}{10001}< \frac{1}{10000}=\left(\frac{1}{100}\right)^2\)

\(\Rightarrow S< \frac{1}{1000}< 0,01\)

Vậy S<0,01

\(\frac{6}{8}< 1\)vì \(\frac{6}{8}< \frac{8}{8}\)

\(\frac{1}{4}< 1\)vì \(\frac{1}{4}< \frac{4}{4}\)

\(\frac{5}{2}>1\)vì \(\frac{5}{2}>\frac{2}{2}\)

\(\frac{7}{3}>1\)vì \(\frac{7}{3}>\frac{3}{3}\)

\(\frac{19}{19}=1\)vì \(\frac{19}{19}=\frac{19}{19}\)