Với mọi k, n Є N+, n ≥ 2 có 1 / (k + 1) + 1 / (k + 2) + ... + 1 / (k + n) < n / (k + 1) 
=> 
1 = 1 
1 / 2 + 1 / 3 < 2 / 2 = 1 
1 / 4 + 1 / 5 + 1 / 6 + 1 / 7 < 4 / 4 = 1 
1 / 8 + ... + 15 < 8 / 8 = 1 
1 / 16 + ... + 1 / 31 < 16 / 16 = 1 
1 / 32 + ... + 1 / 63 < 32 / 32 = 1 
Cộng vế theo vế có 1 + 1 / 2 + ... + 1 / 63 < 6

1+1/2+1/3+1/4+...+1/63=1+(1/2+1/3)+(1/4+1/5+1/6+1/7)+(1/8+1/9+...+1/15)+(1/16+1/17+..,+1/31)+(1/32+1/33+...+1/63)

                                             <1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8+1/8+...+1/8)+(1/16+1/16+...+1/16)+(1/32+1/32+...+1/32)

                                              <1+1+1+1+1+1=6

Ta có:\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..........+\frac{1}{64}\)

=\(1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+.........+\left(\frac{1}{33}+......+\frac{1}{64}\right)\)

\(>1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+...+\left(\frac{1}{64}+\frac{1}{64}+.........+\frac{1}{64}\right)\)

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

=4

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

S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)

(*)Ta có:

1/13<1/12

1/14<1/12

1/15<1/12

=>1/13+1/14+1/15<1/12

(*)Ta lại có:

1/61<1/60

1/62<1/60

1/63<1/60

=>1/61+1/62+1/63<1/60

=>S<1/5+1/12.3+1/60.3

S<1/5+1/4+1/20

S<1/2

S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)

(*)Ta có:

1/13<1/12

1/14<1/12

1/15<1/12

=>1/13+1/14+1/15<1/12

(*)Ta lại có:

1/61<1/60

1/62<1/60

1/63<1/60

=>1/61+1/62+1/63<1/60

=>S<1/5+1/12.3+1/60.3

S<1/5+1/4+1/20

S<1/2

mọi người để ý cau trả lời của bạn Dung và đứa chép bài Trà My nhé vì

1/13<1/12

1/14<1/12

1/15<1/12

nên 1/13+1/14+1/15<1/12(đây là câu trả lời của Dung và đứa chép bài)

bạn sai vì 1/13<1/12

               1/14<1/12

               1/15<1/12

nhưng chưa chắc 1/13+1/14+1/15<1/12(mình chưa tính nhé ^^) và từ đó ta cũng kết luận đc Trà My là 1 đứa chép bài 

Đặt \(A=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{100!}\)

Ta thấy:

\(\dfrac{1}{2!}=\dfrac{1}{1.2};\dfrac{1}{3!}=\dfrac{1}{1.2.3}< \dfrac{1}{2.3};...;\dfrac{1}{100!}=\dfrac{1}{1.2...100}< \dfrac{1}{99.100}\)

Cộng vế với vế ta được:

\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{100}< 1\)

Vậy \(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{100!}< 1\) (Đpcm)

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\dfrac{1}{100!}\)
\(=\left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+\left(\dfrac{1}{3!}-\dfrac{1}{4!}\right)+...+\left(\dfrac{1}{99!}-\dfrac{1}{100!}\right)\)
\(=1-\dfrac{1}{100!}< 1\)

Đặt \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{31}+...+\dfrac{1}{31}\)(có 62 số hạng \(\dfrac{1}{31}\))

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>\dfrac{1}{31}\times62\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\)

\(Vậy\) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\left(đpcm\right)\)

ta có :1+(1/2+1/2+........+1/32)

        1+(1/2*3*4*5......*32)

=>1/2*3*4*....*32<1

vậy 1+1/2+1/3+1/4+........+1/32<3