CMR A = 1/5 + 1/6 + 1/7 + ... + 1/17 < 2

A = 1/5 + 1/6 + 1/7 + ... + 1/17

Vì 1/6 < 1/7 < 1/8 < 1/9 < 1/5 và 1/10 < 1/11 < 1/12 < 1/13 < 1/14 <1/15 < 1/16 < 1/17 < 1/8 nên:

A = (1/5 + 1/6 + 1/7 + 1/8 + 1/9) + (1/10 + 1/11 + 1/12 + 1/13 + 1/14+ 1/15 + 1/16 + 1/17)

A < (1/5 + 1/5 + 1/5 + 1/5 + 1/5) + (1/8 + 1/8 + 1/8 + 1/8 + 1/8+1/8 + 1/8 + 1/8)

A < 1 + 1

A < 2

Vậy: A < 2 (đpcm)

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{50}\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

ko ai trả lời thì để mình

C/M : n/n+1 < n+1/n+2

1 - n/n+1 = 1/n+1

1 - n/n + 2 = 1/n+2

Vì 1/n+1 > 1/n+2 nên n/n+1 < n+1/n+2

1/2 . 3/4 . 5/6 ... 2499/2500 < 1/2 . 2/3 . 3/4 ... 2501/2502

=1/2501 < 1/2500 (1/50) ²

1/50 < 1/49 => A <1/49

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)( đpcm )

a>

\(\frac{1}{2^2}+\frac{1}{100^2}\)=1/4+1/10000

ta có 1/4<1/2(vì 2 đề bài muốn chứng minh tổng đó nhỏ 1 thì chúng ta phải xét xem có bao nhiêu lũy thừa hoặc sht thì ta sẽ lấy 1 : cho số số hạng )

1/100^2<1/2

=>A<1

oke