bạn xét :1/2+1/3+1/4>1 
vậy 1/5+1/6+1/7+1/8...>1 
vậy nó >2 
cách khác. 
tính S62=31*[2*1/2-(62-1)*(-1/6)]>2

163-x=13\5

163-x=2dư3

\(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}\)

\(>\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)\)

\(>\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+...+\frac{1}{8}\right)+\left(\frac{1}{16}+...+\frac{1}{16}\right)\)

\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)

          bạn hãy áp dụng và like nha

Chứng minh rằng: 1 + 1/2 + 1/3 + 1/4 +...+ 1/63 < 6?

trước hết ta cần chứng minh bài toán 1/(k+1)+1/(k+2)+1/(k+3)+…+1/(k+n)<n/(k+1... với n>2,k thuộc N* 
Thật vậy vì k thuộc N*nên ta có 
k+1=k+1=>1/(k+1)= 1/(k+1) 
k+2>k+1=>1/(k+2)<1/(k+1) 
k+3>k+1=>1/(k+3)< 1/(k+1) 
… 
k+n>k+1=>1/(k+n)< 1/(k+1) 
=>1/(k+1)+1/(k+2)+1/(k+3)+…+1/(k+n)< 
1/(k+1)+ 1/(k+1)+…+ 1/(k+1) (có n số 1/(k+1) ) 

=>1/(k+1)+1/(k+2)+1/(k+3)+…+1/(k+n) 
<n/(k+1) 
………………………… 
Áp dụng bài toán trên ta có 
1=1 
1/2+1/3 
=1/(1+1)+1/(1+2) 
<2/(1+1)=2/2=1 
1/4+1/5+1/6+1/7 
=1/(3+1)+1/(3+2)+1/(3+3)+1/(3+4) 
<4/(3+1)=4/4=1 
1 / 8 +1/9 ... +1/15 
=1/(7+1)+1/(7+2)+…+1/(7+8) 
<8/(7+1)=8/8=1 
1/16+1/17+..+1/31 
=1/(15+1)+1/(15+2)+….+1/(15+16) 
<16/(15+1)=16/16=1 
1/32+1/33+…+1/63 
=1/(31=1)+1/(32+1)+…+1/(31+32) 
<32/(31+1)=32/32=1 
=>1 / 2 + 1 / 3+…+1/63<1+1+1+1+1+1 
=>1 / 2 + 1 / 3+…+1/63<6 (đpcm)

sao lại 1,4, là 1/4 chứ

S= (1/2 +1/4+1/6+….1/62)+ (1/ 3+1/5+1/7……+1/63)
ta thấy S1=1/2+1/4+….1/62 có 31 số
1/61 < 1/2, 1/62 < 1/4...... ==> s1 > 1/62+1/62 +….+1/62 (31 số ) = 31/62=1/2
S2= 1/3 +1/5+…+1/63 có 31 số
ta thấy 1/63< 1/3 , 1/63 < 1/5..... ====>S2 > 1/63+1/63…+1/63(31 số)
S2 > 31/ 63 =1/3
S1+s2 > 1/2 +1/3 = 5/6

giúp mình nhé

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}+\dfrac{1}{64}\\ =\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{64}\right)\)

Ta thấy:

\(\dfrac{1}{3}\) lớn hơn \(\dfrac{1}{4}\)

\(\dfrac{1}{5};\dfrac{1}{6};\dfrac{1}{7}\) lớn hơn \(\dfrac{1}{8}\)

\(\dfrac{1}{9};\dfrac{1}{10};...;\dfrac{1}{15}\) lớn hơn \(\dfrac{1}{16}\)

\(\dfrac{1}{17};\dfrac{1}{18};...;\dfrac{1}{31}\) lớn hơn \(\dfrac{1}{32}\)

\(\dfrac{1}{33};\dfrac{1}{34};...;\dfrac{1}{63}\) lớn hơn \(\dfrac{1}{64}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>\dfrac{1}{2}+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{32}+\dfrac{1}{32}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{64}+\dfrac{1}{64}+...+\dfrac{1}{64}\right)\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\\ \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>3\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}>3\)(ĐPCM)

Help me ! Mình sắp phải nộp rồi .

de co sai khong ban? Mk nghi phai la 1/2+1/3+1/4+...+1/63 > 2 chu?

Ta có: A = 1/2+1/3+1/4+...+1/62+1/63+1/64

A = 1+(1/2+1/3+1/4)+(1/5+1/6+1/7+1/8)+(1/9+1/10+...+1/16)+...+(1/17+1/18+....+1/32)+(1/33+1/34+...+1/64)

Ta có: 1/2+1/3+1/4>1/2+1/4+1/4=1

1/5+1/6+1/7+1/8>1/8+1/8+1/8+1/8=1/8.4=1/2

1/9 +1/10+...+1/16>1/16+1/16+...1/16=1/16.8=1/2

1/33+1/34+...+1/64>1/64+1/64+...+1/64=1/64.32=1/2

Vậy A > 4

Xin ai giải hộ cái

xét từng đoạn 1 , 1/2 ,1/2^3 ,1/2^4 ,1/2^5 ,1/2^6 
ta có 
1 = 1 
1/2 + 1/3 < 1/2 + 1/2 = 1 
1/4 + 1/5 + .. + 1/7 < 1/4 +..+ 1/4 = 4/4 = 1 
1/8 + 1/9 + .. + 1/15 < 1/8 + .. + 1/8 = 8/8 = 1 
tương tự 
1/16 +1/17 + .. + 1/31 < 1 
1/32 + 1/33 + .. + 1/63 < 1 
=> cộng lại => B < 6

vi 1/2+1/3+1/4 da lon hon 1 roi ma