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

2S = \(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{19}}\)

=> 2S - S = \(1-\frac{1}{2^{19}}\)

=> S = \(1-\frac{1}{2^{19}}

nen 2S=1+1/2+1/2 mu 2 +....1/2 mu 19

do do 2S-S=1-1/2 mu 20 .vay S=1-1/2 mu 20 <1

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

\(\Rightarrow2S=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{19}}\)

\(\Rightarrow2S-S=\left(1+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}}\right)\)

\(S=1-\frac{2}{2^{20}}\)

\(\Rightarrow S< 1\left(đpcm\right)\)

Ta có :\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)

\(S=\frac{1\cdot2^{19}}{2\cdot2^{19}}+\frac{1\cdot2^{18}}{2^2\cdot2^{18}}+\frac{1\cdot2^{17}}{2^3\cdot2^{17}}+...+\frac{1\cdot2}{2^{19}\cdot2}+\frac{1}{2^{20}}\)

\(S=\frac{2^{19}}{2^{20}}+\frac{2^{18}}{2^{20}}+\frac{2^{17}}{2^{20}}+...+\frac{2}{2^{20}}+\frac{1}{2^{20}}\)

\(S=\frac{2^{19}+2^{18}+2^{17}+...+2^1+1}{2^{20}}\)

\(S=\frac{2^0+2^1+2^2+...+2^{19}}{2^{20}}\)

Xét: Gọi \(N=2^0+2^1+2^2+...+2^{19}\)

\(2\cdot N=2^1\cdot2^2\cdot2^3\cdot...\cdot2^{20}\)

\(2\cdot N-N=\left(2^1+2^2+2^3+...+2^{20}\right)-\left(2^0+2^1+2^2+...+2^{19}\right)\)

\(N=2^{20}-2^0\)

Thay N vào S, ta có :

\(S=\frac{2^{20}-2^0}{2^{20}}\)

\(S=\frac{2^{20}}{2^{20}}-\frac{1}{2^{20}}\)

\(S=1-\frac{1}{2^{20}}\)

Vì \(1-\frac{1}{2^{20}}< 1\Rightarrow S< 1\left(Đpcm\right).\)

Vậy : \(S< 1.\)

Bạn nhân S với 2

Lấy 2S-S=1-1/(2^20)

S=1/(2^20) nên < 2

Cần làm đầy đủ hơn thì bảo mình

Ta có : 1/2 < 1

             1/2^2 < 1/2

              ..............

                  1/2^19 < 1/2^20

Suy ra 1/2+1/2^2+......+1/2^19<1+1/2+1/2^2+......+1/2^20

Suy ra 1/2+1/2^2+.......+1/2^19+1/2^20<1+1/2+1/2^2+.....+1/2^20+1/2^20

Suy ra S<S+1+1/2^20

Suy ra S<S+1+1/2^20<2

Suy ra S<2

giải chi tiết nha!

Ta có: S = 1/ 2 + 1/ 2^2 + 1/ 2^3 + ... + 1/ 2^20

Nên 2S = 1 + 1/2 + 1 / 2^2 + 1/ 2^3 + .... + 1/ 2^19

Do đó 2S - S = 1 - 1/ 2^20 < 1 

Vậy S < 1

2S=\(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)

2S-S=1-\(\frac{1}{2^{20}}\)

S=\(1-\frac{1}{2^{20}}<1\)

S<1

`Answer:`

 \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{31}+\frac{1}{32}\)

a) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}\)

\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)

\(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}>16.\frac{1}{32}=\frac{1}{2}\)

\(\Rightarrow S>\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}\)

b) Ta thấy:

\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}< 3.\frac{1}{3}\)

\(\frac{1}{6}+...+\frac{1}{11}< 6.\frac{1}{6}\)

\(\frac{1}{12}+...+\frac{1}{23}< 12.\frac{1}{12}\)

\(\frac{1}{24}+...+\frac{1}{32}< 9.\frac{1}{24}\)

\(\Rightarrow S< \frac{1}{2}+1+1+1+\frac{9}{24}=\frac{31}{8}< \frac{9}{2}\)

Ta có:

S = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}=1-\dfrac{1}{20}< 1\)

Vậy S<1

Ta có:

S =

Vậy S<1

Ta có

Nên

Do đó 2S - S =