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\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
a) \(A=\frac{7}{10}+\frac{7}{10^2}+\frac{7}{10^3}+...\)
\(A=\frac{777...}{1000...}\)
b) 1/2+1/3+1/4+…+1/63=1/2+(1/3+1/4)+(1/5+1/6+…+1/10)+(1/11+1/12+….+1/20)+(1/21+1/22+….1/63).
Ta thấy:
1/3+1/4>1/4+1/4=1/2
1/5+1/6+…+1/10>5/10=1/2
1/11+1/12+….+1/20>10/20=1/2
Thêm.cái 1/2 sắn có là đủ >2 rồi nhể
1 + 1/2 + 1/3 + ... + 1/62 + 1/63 + 1/64
= 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)
> 1 + 1/2 + 1/4 × 2 + 1/8 × 4 + 1/16 × 8 + 1/32 × 16 + 1/64 × 32
> 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
> 1 + 1/2 × 6
> 1 + 3
> 4
1 + 1/2 + 1/3 + ... + 1/62 + 1/63 + 1/64
= 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)
> 1 + 1/2 + 1/4 × 2 + 1/8 × 4 + 1/16 × 8 + 1/32 × 16 + 1/64 × 32
> 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
> 1 + 1/2 × 6
> 1 + 3
> 4
Ta có:
1/2^2+1/3^2+.....+1/20^2>1/2.2+1/3.4+1/4.5+.....+1/20.21
=1/4+1/3-1/21
=1/4+6/21
=45/84>1/2
Ta có:
1/2^2+1/3^2+..........+1/20^2<1/1.2+1/2.3+.....+1/19.20
=1-1/20
=19/20<1
Ta có 1/3+1/4>1/4+1/4=1/2
Suy ra , 1/2+1/3+1/4>1
* 1/5+1/6+1/7+1/8>1/8+1/8+1/8+1/8=4/8=1/2 (1)
*1/9+1/10+1/11+...+1/17>1/17+1/17+1/17+...+1/17(9 p/s1/7)=9/17 >8.5/17=1/2 (2)
Từ (1) và (2) , suy ra : 1/5+1/6+1/7+...+1/17 > 1/2+1/2 = 1
Vậy: 1/2+1/3+1/4+...+1/17 > 2
Mà 2 < 1/2+1/3+1/4+...+1/17 < 1/2+1/3+1/4+...+1/63
Suy ra : 1/2+1/3+1/4+...+1/63 > 2 ( ĐPCM )