ta có 1/50>1/100

1/51>1/100

1/52>1/100

................

1/99>1/100

suy ra S=1/50+1/51+1/52+..........+1/99>1/100x50=1/2

suy ra S=1/2

Ta có: 

A = 1/2-1/3+1/4-1/5+1/6-1/7+ ..... +1/98-1/99 

=> -A = -1/2+1/3-1/4+1/5-1/6+1/7+ ..... -1/98+1/99 

=> -A = 1/2+1/3+1/4+1/5+ ... +1/98+1/99 - 2.(1/2+1/4+1/6+...+1/98) 

=> -A = 1/2+1/3+1/4+1/5+ ... +1/98+1/99 -(1+1/2+1/3+1/4+...+1/49) 

=> -A = -1+1/50+1/51+1/52+ ... +1/99 


Đặt: B = 1/50+1/51+1/52+ ... +1/99 

=> B = (1/50 +1/51+...+1/59) +(1/60+1/61+...+1/69) +(1/70+1/71+...+1/79) +(1/80+1/81+...+1/89) +(1/90+1/91+...+1/99) 

Do đó: 

10.(1/59)+10.(1/69)+10.(1/79) +10.(1/89)+10.(1/99) < B < 10.(1/50)+10.(1/60)+10.(1/70) +10.(1/80)+10.(1/90) 

=> 10.(1/60)+10.(1/70)+10.(1/80) +10.(1/90)+10.(1/100) < B < 10.(1/50)+10.(1/60)+10.(1/70) +10.(1/80)+10.(1/90) 

=> 1/6 +1/7 +1/8 +1/9 +1/10 < B < 1/5 +1/6 +1/7 +1/8 +1/9 

=> 0,6456 < B < 0,7456 

=> 3/5 < B < 4/5 

=> -2/5 < -1+B < -1/5 

=> -2/5 < -A < -1/5 

=> 1/5 < A <2/5

ta có:1/50>1/100

         1/51>1/100

          ...............

          1/99>1/100

=>S>50*1/100

=>S>1/2(đpcm)

1/50>1/100

1/51>1/100

...................

1/99>1/100

=>S>50*1/100(do từ 1/50 đến 1/99 có 50 số hạng)

=>S>1/2

Ta có S = \(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{74}+\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+...+\frac{1}{99}\)

\(=\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{74}\right)+\left(\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+...+\frac{1}{99}\right)\)

               25 số hạng                                                    25 số hạng

\(>\left(\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}\right)\)

\(=25.\frac{1}{75}+25.\frac{1}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)(ĐPCM)

Vậy S > 1/2

ta có 1/50>1/100    

         1/51>1/100

       ..........

          1/99>1/100

  vậy S>1/100*50=1/2

suy ra S>1/2

S = 1 / 50 + 1 / 51 +...+ 1 / 99 > 1 / 99 + 1 / 99 +...+ 1 / 99 = 50 / 99 > 50 / 100 = 1/2

\(S=\frac{1}{50}+\frac{1}{51}+.....+\frac{1}{99}>\frac{1}{99}+\frac{1}{99}+...+\frac{1}{99}=\frac{50}{99}>\frac{50}{100}=\frac{1}{2}\)

Mỗi phân số trong tổng đã cho đều lớn hơn \(\dfrac{1}{100}\), tất cả có 50 phân số. Vậy

S → \(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{50}{100}=\dfrac{1}{2}\).