\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\\ 2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\\ 2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{19}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{20}}\right)\\ S=1-\dfrac{1}{2^{20}}\\ =>S< 1\\ \)

vậy S bé hơn 1

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

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

=>2S-S=(1+1/2+1/2^2+...+1/2^19)-(1/2+1/2^2+1/2^3+...+1/2^20)

S=1-1/2^20<1

=>S<1

Vậy S<1

100-(1+1/2+1/3+1/4+...+1/100)= (1+1+1+..+1)+(1+1/2+1/3+1/4+...+1/100) = (1-1)+(1-1/2)+(1-/3)+...+(1-1/100)

                                            = 1/2+2/3+3/4+...+99/100 (đpcm)

S = 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰

⇒2S = 1 + 1/2 + 1/2² + ... + 1/2¹⁹

⇒S = 2S - S

= (1 + 1/2 + 1/2² + ... + 1/2¹⁹) - (1/2 + 1/2² + 1/2³ + ... + 1/2²⁰)

= 1 - 1/2²⁰ < 1

Vậy S < 1

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}}

tham khảo

Ta có 

Nên 

Do đó 2S - S = 

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

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

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

\(S=1-\frac{1}{2^{20}}< 1\)-> ĐPCM.